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ELEMENTARY STATICS

t BY THE UEV.

GEORGE RAWLINSON, B.A.

OF EMMANUEL COLLEGE, CAMBRIDGE, Ain> LATE PBOFBBSOB OF THB
APPLIED BCIENCES, ELPHINSTONE COLLEGE, BOMBAY.

EOITED BY THB BEY.

EDWAED STUKGES, M.A.

BEOTOB OF EBNCOTT, OZFOBDSHIBE.

PUBLISHED UNDER THE AUTHORITT OF BER MJJESTTS 8ECRETART
OF STATE FOR INDIA, FOR USE IN THE GOVERNMENT SCH00L8

AND COLLEGES IN INDIA.

MACMILLAN AND CO.

<2Dambt(trQe:

AND 23, HENRIETTA STREET, COVENT QABDW(,

1860-

PEEFACE.

A Treatise on Elementary Statics for the use of
Students in Elphinstone College was in course of prepa-
ration^by my lamented friend, Mr Bawlinson, when deatli*
brought ali bis labours in the cause of education to a sudden
end. What he had written on this sabject was sent to me
with a request that I would prepare it for the press. In
doing so I found a considerable portion of the pian of the
work only sketched out in the manuscript, this I had to
fili in as best I could. I am also responsible for most of
the problems and iUostrations and for the examination papers
which appear in the foUowing pages, I feel also that I am
solely responsible for the faults that will be found in them.
Whatever portion of the work has any merit may be safely
reckoned as coming from the pen of my friend, whatever is
weak and obscure as coming from my own.

It will perhaps be felt that some important points bave
been omitted which ought to have found a place in this
Treatise, and that others have not been considered with
sufficient fulness. The absence of any explanation of the
Principle of Virtual Velocities — a principle of which, I think,

IT PREFACE.

great use mìght be made in an Elementary Treatise on
Statics — and the want of a larger collection of Problems,
and of advice and guidance in their solution, are doubtiess
blemishes. But I bave been in a measure constrained to
brevity. The work is published at the charge of Her
Majesty's Government, and I bave not felt myself justified
in exceeding, more than could be avoided, the space and
expense which, in the estimate previously supplied by the
Publishers, was deemed sufficifnt for printing the manu^
script as it was left by Mr Rawlinson.

EDWARD STURGES.

Kencote Bectobt,

Octdber, 1860.

CONTENTS-

CHAFTm I.
Préliminary notions,

Definition of force. Art. 1. Numérical estimate of force, 2. Unita
of weìght and length, 3. How to represent forces by straight
lineS) '4. The use of the signs + and ~ before symbols re-
presenting forces, 5. Axìohìs, 6. Examination on Chap. I. 1 — 9

CHAPTBR IL

On the Composition and ResoluHon of Forces cutting ai one point,

Any namber of forces acting on a point in any directions must
hare a resultant, Art. 1. To find the resultant of any namber
of forces acting on a particle along the same line, 2, 3. If
any number of forces act npon a particle and keep it at rest,
any one of them is equal and opposite to the resultant of ali
. the rest, 7. The direction of the resultant of two equal
forces acting on a particle in different directions, 8. The
enunciation of the parallelogram of forces, 9. Examples on
the composition of two forces, 10. The trìgonometrical for-
muke fot finding the magnitudo and direction of the result-

VI CONTENTS.

PAOB

ant of two forces, 11. On the resolution of a force into a
pair of component forces, 12. Besolution of a force when the
dìrectìons of the component forces are gìven, 13. Examples,
14y 15, 16. Examination on Ghap. II 10 — 28

CHAPTBR III.

On the Oomposition and Resolution ofForcei ^xcting on a point,

Proof of the parallelogram of forces, Art. 1. The triangle of
forces, 2, 3. The polygon of forces, 4, 5. Extension of the
parallelogram of lòrces by geometry, 6. To find the magni-
tudo and direction of the resultant of any number of forces
actlng on a point, 7. To find the conditions of equilibrium
of any number of forces acting on a particle, 8. Problems,
9. Examination on Ghap. Ili 29-- 47

CHAPTER IV.
On the Composition and BesoliUion qf Forces acting on a rigid body.

How theinrestigation of forces in equilibrium acting on a particle
is extended to a rigid body. Art. 1. The resultant of two
parallel forces, 2. Definition of Gentre of Gravity. Gentre
of Grayity of a uniform beam, 3. Two equal and parallel
forces acting in opposite directions can hare no single re-
sultant. Definition of a couple, 4. Definition of moment
of a force about a point, 6. How any number of forces act-
ing on a rigid body at different points in the same piane may
be supposed to act ali at one point by the ìntroduction of
couples, 6. The effect of a couple is not altered by tumiiig
its arm through any angle about one extremity in its own
piane, 7. Two couples whose arms bave a common extremity
and whose moments are equal bave the same statical effect,
8. To find the resultant of any number of couples whose
arma hare a common extremity, 9. To find the conditions
of equilibrium of any number of forces acting on a rigid body,
10. Problems, 11. Examination on Ghap. lY. . . 48—69

CONTENTS. vii

CHAPTER V.

On the Compoaition and Reaolution cf Force* acting on a system

o/two or more rigid bodies,

• PAOB

Action and reaction, Art. 1. Contact of smooth snrfaces, 2. Pro-
blems, 3. Contact of rongh surfaces, 4. Problema, 5. Ex-
amination on Chap. Y 70 — 81

CHAPTER VI.
On the Centre cfOravity and Equilvbrium,

,Property of the centre of gravity, Art. 1. The centre of gravity
may sometimes be determìned by inspection, 2. Given the
centres of gravity of two parts of a body to find the centre
of gravity of the whole, 3. To find the centre of gravity of
a system of material particles^ 4. The centre of gravity of
a trìangle, 5. Of a trìangular pyramid, 6. Of a cene, ib,
Of the surface of a cono, 7. Problems, 8, 9, 10. On equi-
librìum, 11. To determine the centre of gravity of a body
by experìment, 12. Stable and tinstable equìlibrìum, 13*
Examination on Chap. YI. f • • • • • 82 — 99

CHAPTER VII.
On Machines,

Preliminary remarks, Art. 1.

The Lever. The three different kinds, 2. Examples, 3. The
modulus, 4. The modulus obtained geometrìcally, 6. The
common bsdance, the requisites for a good one ; the measore
of sensibility, 6. The common steelyard, 7* The Danish
steelyard, 8. The compomid lever, 10.

The Wheel and Axle. Descrìptionof,ll. The modulai^ 12. The
vdndlass» 13. The capstm, 14, Wheels and axles in combi-»
nation^ 15. Toothed wheels, 16,

TIU

CONTENTS.

TAQM

The Pulley. Descrìption of, 17. The angle moveable pulley, 18.

The first system of pulleys, 19. The second system, 20.

White's pulley, 21. The thìrd system of pulleys, 22. Be-

marks on some of the catises of inaccuracy in onr resulta.
The inclined piane, descrìption of, and modulus, 24.
The Screw, descrìption of, and modulus, 26.
The Wedge, descrìption of, and modulus, 26.
Machines with frìction. The Lever^ 27. The wheel and azle, 28.

The inclined piane, 29. The screw, 30. The wedge, 31.

Examination Papers on Chap. VII 100 — 143

Problems 143 — 160

* ^ — - V ^

ELEMENTARI STATICS,

CHAPTER L

PRELIMINARY NOTIONS.

1. Stàtics is the scìence which ìnvestigates the con-
ditiòns which must hold among any number of forces which
keep a body at rest, We define force thus :

. DsF. Farce is the cause which praduceè w tenda io pro^
duce chaohge in the state ofrest or mation ofa body.

In Statica we are concemed only with those forces which
iend to produce the change. When sùch a change is actually
prodticed the treatment of the subject falls under another
Dranch of Mechanics, called Dynamics.

' 2, The first thing to be done in order to bring force'
under Mathematica! treatment is to find some ineÀns by
T^hich the intensity of different forces may be numerically
lepresented.

It will be guessed from the definition of force, that ali onr
knowledge of this subtle agent is derived from the effect we
see it produce rather than from any direct knowledge of ìt.
Thus when we lift a weight with the hand, we are conscioud
of s^ force exerted on the weight by the mu^cles of the arm.
We know such a force can be called forth at pleasure by a-
connection existing between the muscles and the brain, Wt
when we have thus traced the source of this power from one^
seat to another, and are asked " What is this force?" we can

B.S. 1

2 PBELIMINARY NOTIONS.

only anBwer, ^'It ìs samething which enables us to lift a
certain weight."

Accordingly we estimate force by the effect it produces or
is capable of producing.

To arrive at 9,,mimerioal estimate of a force^ we mnst fìirtlier
fix, by common consent, upon some standard force which
producing a known effect may be taken as a unit of force and
De represented by the figure 1, a force producing twice that
effect will be represented by 2, and so on.

It is usuai to take ap pur unit of force, that force which
will sustain llb. weight when acting vertically upwards, a
force which would sustain 2lbs. will then be represented by
the figure 2, a force which will sustain 3lbs. by the figure 3,
and generally a force which will sustain Plbs, will be re-
presented by P. , ->

This then enables us to represent and compare forcés
numerically.

. 3. Force is alwavs meaaured numerically upon this prin-
cipìe, that is, by the weight it will sustain when acting
vertically u^wards ; althoiigh the unit of weight varies with

f)lace and circumstance. For instance, in dealing with very
arge forces, we might speak of a force of so many tons in-
stead of pounds ; and an inhabitant of Bombay may speak
of a force of so many seers.

Now fiirther, to clear up our ideas on this subject, let us
inquire what is meant by a unit of weight; a pound for
iinstance. What is a pound? The question seems easy
enough. We have in our eye a piece of lead or iron of a
certain size. We know by experience pretty well what a
pound of iron is, or a pound of sugar ; but we want to know
what security we have for a pound being the same weight
ali the world over. If it were suspected that ali the pound
weights in India had by wear or other cause lost some of
their originai weight, how could we get back to our
standard?

To shew some of the difficulties of fixing upon a standard
of comparison, suppose, for a moment^ we nad agreed to say

^SELIHINART NOTIONSl 3

that a ponnd should be a cubie incli of iron. Now there are
two objections fatai to such a selection of a standard. In the
first place metal expands by heat, and consequently our pound
would be more than a cubie inch in hot weather, iess in cold^
And secondly, different kinds of iron bave different densities,
so that a cubie incb of one kind would weigh more or Iess
than a cubie inch of another,

We must therefore search for a standard in some substance
which is or can be made free from these objections. Now
pure distilled water at a given temperature is always of the
same weight per cubie inch. And accordingly a pound is
fixed by Act of Parliament to be the weight of 22*185 cubie
inches of pure distilled water at the temperature of 62^
Fahrenheit.

22'185 cubie inches of distilled water at 60® being an in-
variable quantity, the pound so estimated is invariable also.

The method of measuring water with sufficient accuracy to
obtain 22'185 inches, forms no part of our present inquiry.
It is sufficient to say that it can be done. It will be seen
that our unit of weight involves the idea of a unit of length,
for we say it is so many inches of a certain substance.

What then is an inch? Here again we must seek for
some standard of comparison which neither beat nor want of
uniformity in the material, nor any other cause will affect.
The Act of Parliament defines an inch to be the 39'1393* part
of the length of a simple pendulum vibrating seconds, in the
latitude of London,. at the level of the sea.

This again involves the idea of a unit of time, a second,
but this is obtained directly from the motion of the Sun
round the Earth.

4. To return now to our subject, let us conslder what
this numerical estimate of force which we bave arrived at will
do for us; that is, what Information is conveyed to us
when we talk of a force or pressure of lOlbs. Clearly only this,
that a force of a certain intensity (namely lOlbs.) is acting
somewhere in a body in some direction. And a very little
thought will shew us that we must bave this somewhere

1—2

nuSUMINABT NOTIQKa

Cmted oat as well ^ the dtreciioH in wbidb the force acts,
fere we can form an idea of the effect of the force. In other
woida we muat know not ouly the irUensity of the foroCi but
ita paint of c^fpltoation and its directhnu

Fot instance, if I teli jou a body À Ijing on the table is

Q 1

aeied npon by a force of lOlbs., yon will not know whether
this force is pressing the body down, or lifting it np, or push-
ing it along, or actìng in any other direction. But if 1 say
the force acts on the body at the point g^, and in the direction
represented by the line (8j, you can now form some con-
ception of the character of this force, and will be prepared to
reason on the effect it will produce on the body.

There are, then, three indispensable things which must be
^ven US before we can say a force is hnown^ namely:

1. Its magnitude»

2. Its direction.

d. Its point of application.

Now there is a way of representinff forces by which thcse
three data may ali be presented to me eye at once, namely,
by straight lines.

We will explain this by an example.

Ex. Suppose a force of 5lbs., inclined at 30* to the
horizon, to oc acting at the centre of a horizontal rod.

Let j5(7 be the rod, A the centre, draw at ^ a straight
line making an angle 30^ with the horizon, take a portion
of this line AP containing 5 units of length, Le. as many
units of length as there are units of force in the given force«
Then we say AP represents the given force in every neces«»
eaiy particular,

FRELtMIKÀItT N0TI0N6*

In magnitude, "by the number of units in ita length.

In direction, by its inclination to the horizon, yìz. 30°.

In point of application, vi2, at A the centre of the rod.

It ìs evident that any force mar in this mann^ be fdlly
represented by a Btraight line. Whenever, tìien, in future
we say, Let such and such a straight line represent such and
such a force, the student must remember that each unit
of length of the line represents a unit of force, the incli-
nation of the line to some known line gives us the direction
of the force, and a giyen point in the line gives os its point
of application.

5. The symbol P applied to a force, we bave already
seen, means that the force is of such intensity that it wiu
support a weight containing F times a certùn chosen unit
of weight.

The mode of representing forces by straight lines, suggests
to US a way in which the meaning of tàis symbol may be
further extended.

Every one acquainted with Trigonometry is familiar with
the &ct that the signs + and — are used to indicate whether
a line is measured in one direction or the opposite with regard
to a fixed point.

Thus, if -4 be a fixed point in the line JffAB and AB be
taken equal to AB' equal to a.

£ PBELIMINART NOTIONS*:

If AB 18 called + o, then Aff will be represented by — a.
We adopt a similar convention with regard to forces.

Thns if AB be taken of such a length that it represents
in magnitude and direction a force Pacting on a particle A,
firom ^ towards Bx the symbol — P will represent an equal
force acting in exactly an opjposite direction, geometrically
represented by the line AB\ This meaning of the algebraical
sisns affixed to symbols representine forces must be bome
in mind by the student. ^

6. It would be consistent with the arrangement of cut
subject, to give here a list of definitions of terms nsed in
this Science. But we prefer inserting them from time to
time in the progress of the work, as the necessity of nsing
the terms themselves arises. Some definitions, indeed, are
ìncluded in the explanatory matter, and the student will
be able to draw them thence without their being stated
more explicitly. We have already defined farce. We add
here two other definitions.

Def. a partide^ or material particle^ is an infinitely
small portion of matter of insensible weight, incapable of
dirision or compression.

Dep. Where two or more forces acting on a particle or
system of partidea, produce together no statical elfect what-
over, they are said to be in Equilibrium,

We now cali the student's attention to certain truisms or
axioms. These axioms, although not capable of auy direct
proof, are confirmed by experiment as well as by the unvary ing
correctness of ali results based on the supposition of their
truth*

AxiOM 1. If two equoblfarces act upon a particle in oppo--
site directianSy that point will remain at rest.

^ ' It is «vident firom this Axiom together with Axiom 3, that
two equal forces acting in opposite directions at the ex-
tremities of a straight string will neutralize one another, or
be in equilibrium; and the same will be trae if the string
instead of being straight, is passed over one or more puUeys,

teELIMINÀBy HOriONSi 7

attomiiig the àbienoe of ali frictiòn. &ence this Axiom jà
sotnetìmes spoken of as the Frinciple of the Fulley.

The next axiom is the converse of this«

Axiom 2. Ifapartide remain at ran^^^fR the forces which
act upon it may be rediused to two ^(M^ <ind apposite jbrcea.

• Axiom 3. If a Jbrce te 4ii^lied at any point of a system e/
particles rtgtdly /jptm^rfed together, its statical effect mll he
the same ^jk ie removed to any other point in its direction
ryiiàif iS0mnected with the system ofpoints on which it acts.

This Axiom is the Physical Frinciple of the transmission
of Force,

The next axiom is the converse of this.

AxìOM 4. J^ a force act at any point of a system of
particles rigidly connected together, and hetng removed to any
other point rigidly connected with the system^ he found to
produce the same statical effect^ the linejoining the two points
ù the line of action ofthe^ce.

■

Axiom 5. Ifany system of forces in equilibrium be added
or superimposed on another system also in equilibrium^ the
iohole will be in equilibrium.

This is called the principle of superposition of forces, and
more will be said of it at a mtnre time.

7. It is only by the diligent use of illustrations and ex-
amples that the leamer can make himself familiar with that
process of reasoning, upon his readiness in using which,
nis ultimate success in physical science must depend. We
shall endeavour therefore to illustrate our subject from time
to time with examples and other matters calculated to en-
courage this process of reasoning, This part of the work
will repay the learner for a carefal and dose consideration*

The following is an illustration of Axiom 3*

Suppose a rope be fastened to a nail in a ceiling. Let Q,
be the point of suspension, OC the rope.

; NcFw " if a force be .lujting in the direction of the *
rope, the , axiom tells ub that it wiU be immaterial
at what poìnt in the rope we suppose this force to
act, whether at a point -4 or J9 or 0. Suppose the
force were a weight. of lOlbs., and we hung thÌ9 at
first at A with a rope pf length OA, and then at B,
increasing the length of rope to OB. It might be
. objected that the effect in the two cases would not be
the same, for, in the second case we bave the weight ^^
òf an additional piece of rope AB. This is true, but
observe, the effect of the jvrce 10 IbjS, is the same in
both cases. The fact that the effect of the force at
rB' only exceed^ the effect of the force at Ay by the
weight of the portion of rope AB, sufficiently proves ^
that the force of JO Ibs. does produce, as far as itself j.
is concemedy the- same effect at A and B.

In estimating the effect of forces, the student ghould lèam
to separate, when necessary, each force from its eonnectioisi
with ali other9, so as to be able to cwsider it separately.

Examination (m Chapter I.

1, Distinguìsh betweeu Statica and Dynamics,

2. What would you understand by a force 10, or a force
Pin the absence of otber Information respecting sucb fiMrcee?

(Ans. A force is measured numerìcaUy by the weight it
would sustain if acting vertically upwards. The most common
lujit of weight employed in this case is 1 Ib. Hence, in the
absence of other Information, I should understand by a force
10 or a force P, a force of such intensity that it would
siipport 10 Ibs, or Plbs. respectively, if applied to the weight
in tne manner stated.)

3; How Ì8 a permanent standard of measurement in
length secured?

4. Define the direction afa force»

(Ans. If we suppose a force to act on a particle which
is perfectly free to move, and not acted on by any oth^r

PRELIMtNART I^TIONS, 9

force, the direction in whìch that particle would begin to
move is the direction ofthejbrce.)

5. State clearlj how a straight line may be made to
represent the magnitudo of a force^

6. If 5 represent a force of 5 Ibs, pnlling a body vertically
npwards from the ground, wliat does ^ 5 represent ?

7. Enunciate the principle of the puUey,

8. Enunciate the principle of thè superpositìon of forces» '

( 10 )

CHAPTER II.

ON THE COMPOSmON AND BESOLUTION OF FOBCES ACTING

AT ONE POINT.

1. A LITTLE conslderation wìll satàA ìIm^ jùdeat that
any nandicr of ìhobb actiiig im m poEiticie in any direction
may be represented by a single force which will be equi-
valent to the whole system in every statical particular. For
suppose such a system did not Keep the particle at rest,
it is evident the particle wonld begin to move in a certain
direction. A single force of the necessary magnitude and
actìng in a direction exactly opposite to mat in which the
particle will begin to move, womd keep it at rest. Cali this
force R, then R is in equilibrium with the whole system of
forces; but if so, it is evident (Axiom 2) that the system
itself is equivalent to a single force of the same mamiitude
as jB, acting in the direction in which the particle will begin
to move. R is termed the reaultant ofihe system of forces.

The method of determining the resultant of different
Systems of forces, forms a most miportant branch of Statics.

This process is called the Composition of jbrces. The
opposite process, of deducing froni a single force an equivalent
system, is termed resolution.

We proceed in this and the next chapter to shew how
to determine the resultant in ali cases where two or more
forces act on a particle.

The student will notice that strictly speaking the case
of two or more forces acting on a particle does not fall

ON THE COMPOSITION AND BESOLUTION OF FOBCES. 11

into the snbiect '^ Statica*' at ali, nnless the forces hapoen to
be in equilibrìiim. But ali such cases maj be tnmed ìnto
Statical problema, bj supposing such a force applied to the
system as will keep the whole at rest. This force, it must
De remembered, with the sign changed is the resoltant of
which we bave been speaking.

2. To fini the reaultant of two forces acting on a paritele
along tì^e same line.

If a force of Plbs. acts npon a particle in a certain direction^
and another force of Q Ibs. acts npon the particle in the same
direction, it is evident that the two forces together will pro-
duce or tend to produce the same effect as a single force of
(P+ Q) Ibs. acting in the direction of the two lorces. Or,
if li be the numencal vaine of the equivalent force in Ibs.

p+Q^B...:. (1).

If Pact in one direction and Q in the opposite, and Pbe
the greater force, we shall bave

P-Q^B i (2).

If P he less than Q^ B will be negative, which will
indicate that the particle instead of being pulled forward will
be pushed backward,

By observing the arrangement with regard to the alge-
braical si^s prefixed to symbols representing forces, pointed
cut in Chap. i. Art. 5, the formiuae {!) and (2) will both
be included in the single expression,

P+Q^B.

3. To find the resuttant of any number of forces acting on
a particle along the same straight Une.

This is the preceding problem made c^uite general: let
Pj, P„ P„ &c. be any number of forces actmg upon a particle
along the same straight line, P^, P„ &c. being positive or
negative as the case may be, then we bave

P, + P, + P3+, =5,

12 OK THE COHPOBITION AKD BESOLUTION OF

a £}rniulà which iriU give os a single equivalent force iù
eveiy caae.

Observatton. It wiU be remarked that P^, jP^, P„ &c, and
— jB are a system of forces in equilibrium, xlierefore the
above is a trae statical problem. The condition of equi-
librium being expressed by the equation

Ex. 1, Suppose a particle pnlled in a ceitain direction
by two forces of ^ and &lbs. xespectively, and held back in
the opposite direction by a force of 2 Ibs. : what single force
will produce the same effect on the pajrticle as these tbxee?

Here referring to the general formula of the preceding
proposition, we nave

or, the single equivalent force will be one of Glbs,, acting
in ihe same direction as the forces of 3 and 5.

Ex* 2. Find the condition that the forces P., P^*^. P^ may
be in equiUbriunu

In this case B must equàl 0, therefòre the required con-
dition is

Pj + P, + &c. + P. = 0.

Observatton. This last example is a type of a large class
of Statical Problems. Whenever we bave an equation, or
a number of equations, firom which we may determine the
resultant of a system of forces, by supposing that resultant
to be 0, every such equation will give us a condition which
must hold among the forces in order that there may be
equiUbriura.

•

4. Hitherto we have been concemed with forces acting
along the same straight line; The next proposition, in
order, would be to fina the resultant of two forces acting on
a particle in any directions, This is in fact the great prò-

F0SCB8 ACnNa AT 0N£ POINT« 13

S~>sitioii npon trhich the whole seience of Statica is foonded.
efore wq proceéd to it we have aoine preliminaiy matter
to present to the stodent.

6, It has alreadjr been shewn, in Art 1, of thìs Ctapfer,
that any ^amb^r of forces acting on a particle must have a
resultant; this perhaps -mll appear plamer in the following
manner.

Let P and Q be two forces acting in the particle A in
different directions. Draw -4P, A Q representing the forces in
magnitude and direction.

Now consider what tendency to motion these two forces
produce on A, We observe

, Ist. That the effect of the
force P is to make A tend to
move in direction AP.

2ndlr. That the effect of the
force Q is to.make A tend to
move in direction .4^.

drdlj. Both these forces act
at once, but it is evident the
particle cannot move in both
directions at once. Hereafter
we shall.be able to ascertam
both, the direction in which A
would move under these clr-
cumstances, and also the effect in magnitude of the two forces
upon A* At present we onlj wish to point out that A miist
evidenti/ have a tendencj to move in some direction lying
between.4P and ^Q, '.4^ suppose, therefore a single forqe,
of a certain magnitude, acting in direction AH, would prodiM^e
the same effect on A bs P and Q together. Such a force is
the resultant of P and Q.

6. In the same way it mav be shewn that if any number
òf forces act on a point, these fcrceà together produce the same
effect as a single force R suppose. For if P^jP, ... P, be n
forces acting upon a point -4, any two of them as P^, P^ may

14 ON THS COMPOBrriOli AND BESOLUTION OF

be lemoved and replaoed br a sinele foroe^ B.; similarlj jB
and P, mar be remoTed and replaoed bj a single foioe JB, ana
so on, unti! there remains only a single force È.

B is said to be the lesnltant of Pp P, ••• P«,

7. It foUows immediately from the foregoing that if any
number of force» act upon a paritele and Jceep it ai reat^ any
one ofthem ù equal anì apposite to the resubant ofall the reaU

Let Pj, P, ... P. be six forces
acting on a point A. Then anj
one of them as P^ is equal and
opposite to tàe resoltant of ali the
rest.

By Art 6, it is evident that a
Pj, P^ ... P^ may be removed and
replaoed hj a single force iZ, act-
ing in some direction AB suppose.
Let this be dono. Then we nave
two forces acting in directions ''^

AP^ and AB on tue particle A and keepin^ it at rest. Hence
it follows that P^ and B^ must be equal ana opposite.

Therefore anj one force as P, is equal and opposite to the
resultant of ali the other forces P^, P, ... P^.

8. We maj now proceed tp determine the direction o{ the
resultant in the particular case of two equal forces P and P
acting upon a particle Ay in different directions.

The consideration of this case will shew us that as A can
bave no more tendencjr to move in the direction of one force
than that of the other, masmuch as the forces are equal, there
can be no reason whj the direction of the resultant should be
more inclined to one force than to the other. Therefore the
direction of the resultant will bisect the angle between the
forces.

It will be notioed bere that the direction of the resultant
is that of the diagonal of the parallelogram, constructed upon
the lines representing the two forces*

FOBCES ACTINa AT ONE FOINT* 15

9. The proposition whìch enables us to detennine the
direction and magnitude of the resultant of any two forces
acting on a particle in different directions is called the
" Parallelogram of Forces,"

The student must thoroughlj imderstand the principles
ìnvolved in this important proposition before he can make
anj progress in the science«

We prooeed now to enunciate iU

The Parallelogram of Forces*

If two Jbrces acting upon a paritele in different directions he
re^esented in magnitude and direction hy two straiaht lines
drawnfrom the particle^ then the diagonal of the parallelogram
constructed upon these two straight lines will represent the
resultant of me. forces in magnitude and direction^

We postpone the proof of this proposition for the present.

The following illustration will make the meaning of the
enunciation quite plaìn.

Let Pand Q be two forces acting on the particle A in
directions APy A Q respectively .

Take AP, AQ, each as many
units of length as there are units

in the forces P and Q respec- rr ^

tively, / \ /

Le. ifPbelOlbs. / \ /

and Q be 7lbs. / \ /

..M.................

and one inch be chosen as our
unit of length,

take -4P=t 10 inches,

AQ=^ 7 inchesj

ON THB CX)UP08mOH AND lUiaOLUTIDN OF

then the linea AFj AQtepteaejot the foices Paad Q Iti
Ist. Magnitude.
2nd.- Direction*
8rd. Point of application*

Complete the parallelogram APRQ, and draw the diagonal
AB.

Then the proposition dtates that the diagónal AR irìll
represent the resultant of P and Q in the above three par-
tìcnlarSy viz. in magnitude, in direction, and eyidentlj also
in point of application, P and Q might thercfore be removed,
ana the single force, represented hj the line AB, substitated
in their steiid. ^

10. "Novr assuming the trath of the '^ parallelogram of
forces,'* wegive here some examples whioh will serve to make
the stodent familiar with ita use and principio.

Ex. 1. Two forces of 4 and Slbs. respectively, act on a
pajrticle at ri^ht angles to one another. Find the direction
and magnitude of their resultante

Let A be the particle,

APy AQ the direction of the
forcés.

Let — th of an inch be our

unit of measure.
Tske AB ~th&,

A C TTrths of an inch.
10

Then AB, A G represent the forces respectively.

Complete the parallelogram ABDC. Draw AD.

Then, by our assumption, AD represents the resultant in
magnitude and direction.

^:p

FOBCBS ACTINO AT ONE POINT; 17

If ^2? be measured it will be found to be -rTrth of an inch.

And if the angle BAD be measur^ it will be found to be
about 36°. bV.

Therefore, the resultant is a force of 5lbs. making an
angle of 36°. 51' with the larger force.

This is solving the problem by actual admeasnrement.
The result would be obtained analytically thus :

= 4» + 3'
= 25j
.-. AD=^b.

And sine of angle BAD = -jjz

^3
"5

= •6,

which on reference to the table of naturai sines we find to
be 36°. 51'.

Ex. 2. Two forces of 5 and lOlbs. respectively, whose
directions are inclined at an angle of 120° act upon a particle ;
find the direction and magnitude of their resultant.

Let A be the particle.

Make the angle BAC-=^ 120°.

Take AB containing 5 units of length,

AG 10

Complete the parallelogram A CDB.
B. S. 2

18 ON THB COMPOSmON AND RESOLUTION OF

Then by actual measurement we shall find

AD = 8*65 . . . unita of length,

angle BAD = 90^

or the resultant will be a force of 8*65 ...Ibs. making an
angle 90*^ with the force 5.

Or the problem may be solved geometrically by Euclid;
Book II. Prop. 13,

or by trlgonometry thns,

Aiy = AB^ + BD'^2AB.BD.cosABD,

cos ABD =: cos (180 - 120) = cos 60 = ^

.-. ^li>»=5'+10'-2.5,10 i

= 125 - 50
=^75;

.'. AD = 5»/3 =8-65

(1).

FORCES ACTING AT ONE POINT. 19

Again,

sin BAD BD 10

sin^.B2> AD 5^S V3'
and sin ABD = sin 120" = ^ :

/. sin BAD = 1 = sin 90" ;

/. BAD=^9Qr (2).

(1) gives US the magnitude,

(2) the direction of the resultant.

11. Referring back to the preceding figure, we have the
well-known trigonometrical formula which we have already
used,

AD'=^AB' + BD'-2AB.BDco&ABD.

Therefore if P and Q be two forces acting on a particle and
inclined to one another at an angle a, and if B be the resul-
tant, the angle it makes with P, the above formula be-
comes

B!' = P'+Q^ + 2PQcosa;

we have also

. /, G .
sm a = -^ sin a ;

two general expressions which wiU determine the magnitude
and direction oi the resultant in ali problems of this class.

12. The preceding examples relate to the "Composition
of forces." Assuming for the present the truth of the parai-
lelogram of forces, we proceed now to consider the converse
case, namely, the "resolution" of a single force into two
others.

Let jB be a force acting upon the particle A, represented in
direction by the line AB. Take AD of such length that it
represents this force in magnitude.

2—2

20 ON THE OOMPOSITION ANP EESOLUTION OP

Draw any line AB of any length, join BD. Complete the
parallelogram ABJDC.

Then it is evident that AD is
the resultant of the two forces
which are represented in magni-
tude and direction by the lines AB, -^
AC, respectively.

The single force R may, there-
fore, be removed, and these two
substituted in their stead.

The student will observe that
while a single pair of forces must,

by the parallelogram of forces, bave one and only one
resultant, in the converse case a single force may be resolved
into an infinite number of pairs of forces, any one of which
will be equivalent to the single force.

Thus in the preceding figure, we may draw any other line
instead oi AB, as AE, of any length, join ED, complete the
parallelogram AEDF. Then, it is evident, AD is the re-
sultant of the forces represented by the lines AE, AF; and
so on, we may shew that there are an infinite number of
pairs of forces, any one of which has the force B for a
resultant.

In the two new forces, P and Q suppose, into which we
propose to resolve any given force B acting on a particle,
there are in fact four elements concemed, viz. the magnitude
and direction of P, and the magnitude and direction of Q»
Any two of these may be chosen arbitrarily, except in the case
where we choose the two magnitudes, when it is evident
our choice will be subject to the condition that the magnitudes
of P and Q must be such that any two of the forces P, Q, B
are together greater than the third.

Hence, four classes of problems arlse from this case of the
.resolution of a force. Three of these are given in Example 6,
in the examination paper at the end of this chapter, the
fourth is treated in the foUowing article.

FORCES ACTING AT ONE POINT.

13. A force B acts on a particle A, it is reqmred to re-
solve it into two others F and Q, which shall act in given
directions.

Let Ali be the direction of the force B.

Take AD representing the force
i? in magnitudi

Through A draw AP, A Q pa-
rallel to the given directions of P
and Q respectively.

Draw DB, DG parallel to AP,
A Q respectivelj.

Then AB, AG will evidently
represent the required forces in
magnitude and direction.

The foUowing is a form of the above proposition which is
constantly occurring in statical problems.

To fini the effect of a force in a given direction.

Let ABP, in preceding figure, be drawn from the point of
application parallel to the required direction. Then we want
to find what tendency the force B has to pulì the particle
along the line ABP,

From D draw DB perpendicular to ABP, and complete the
parallelogram ABDG.

Then the forces represented by the lines AB, AG, eoe
equivalent to the force B.

But the force AC being perpendicular to AB can produce
no effect at ali in that direction.

Therèfore the force AB represents the whole effect of B in
a direction parallel to ABP.

ON THE COMPOSITION AND BBSOLUTION OF

In ali fìiture investigations, then, if AB represent a force
in magnitudo and direction, and we want to fìnd the effect of
that force along any line parallel to AM^ draw BN perpen-
dicular to Ami «^d AN will represent the portion oi the
force required.

14. The foUowing are examples of the application of the
principle explained in the preceding article.

Ex. 1. AB is a stick fastened perpendicularly into the
ground, DA is a force pushing on the top of the stick, whoBc
magnitude is 5lbs. and whose direction makes an angle of 30^
with the horizon. Find the tendency of this force to push
the stick on one side, and also its tendency to push the stick
into the ground.

Take DA representing the force in magnitudo and direction.

Produce length of stick in direction AK

Draw DH perpendicular to AE.

Then DE will evidently represent the
magnitudo of the forco which pushes the
stick on one side.

And EA will represent the magnitudo
of the force which pushes it into the
ground.

And since

DA = 5, and angle DAE= 60^

DE=^ DA sin DAE=: 5 sin 60 = 4'33 ... "?

and EA = DA cos 60^ = 2-5.

Hence the single force of 5lbs. acting at an angle of 30* to
the horizon is equivalent to two others, one pushing the stick
downwards with a force of 2*5 Ibs., the other on one side with
a force of 4*33. ... Ibs.

In the foUowing examplo the puUeys are supposed to re-
quire no appreciable force to make them run round, the effect
of them will, therefore, be to transmit the force oxorted by the

FORCES ACrriNG AT ONE POINT.

weights alonff llie whole length of the string without alteration.
Thìs must always be understood to be our meaning when we
speak of ptdleys as smooth.

Ex. 2. A and B are smooth puUeys over whìch passes a
string PACBQ. At each extremity of the string a weight
of lOlbs. is attached, and at the point G another weight of
lOlbs. is attached by a string. Fmd the position the string
wiU take when the system is in equUibrium.

Take (7r, Gm^ Gn^ each representing a force of lOlbs.

It is evìdent that at G we bave ihree forces acting, one
vertically downwards, and one along each of the directions
GB, GA.

These forces are represented in magnitude and direction by
the lines (7r, Cw, Gn respectively.

Complete the parallelogram GnSm*

Draw the diagonal G8.

Then G8 represents in magnitude and direction ihe re-
sultant of the forces represented by the lines Gm and Gn.

But (Art. 7) this resultant must be equal and opposite

ON THE COMPOSITION AND BESOLUTION OF

to the force represented by the line Or. Therefore C8 ì&
yertical, and eqoal lOlbs.

And the triangle CnS is eqnilateral, each of its sides being
equal to 10.

Therefore the angle BCn « 60^

and similarly the angle 8Cm = 60*.

Hence the portions of the string (7-4, CB make equal
angles with the vertical and the whole angle ACB= 120®,
which determines the position of equilibrium.

15. When a force is conducted along a string, as in the
preceding problem, the force so communicated is spoken
of as the tenston of the string. When the string is without
weight, flexible and inextensible, the tension is the same at
every point of its length. This is a matter of experiment.
It is also evident from the fact that such a string is mcapable
of exerting any force in a direction perpendicular to its length,
so that the whole force will continue to.be transmitted without
alteration along the string.

16. ExAMPLE. CBW ìa A string
hangin^ from a peg C to which it is
fastened, and supporting a weight W.
Another string fastened at B draws
the string with a force F, acting hori-
zontally, out of the vertical, into the
position CB'W; find the position of
equilibrium.

Let the figure represent the system
in a position of equilibrium. Let the
angle BCB* be 0. If we determine
the position of equilibrium will be
known. The point B is kept in equi-
librium by three forces jP, tV and the tension of the string
along BG. Produce GB' in direction B'b.

Take B'w representing W in magnitude.
Praw vm perpendicular to GBb.

FOBCES ACTING AT ONE POINT. 25

Then ton, ffn are the resolved parts of W perpendicular to
and along CB'b respectively.

And ton = wB sin wlfn = W sin 6.

And similarly the resolved part of jP perpendicular to CSh
(since WBF is a right angle)

= jPcos^.

And these forces perpendicular to GBh must be equal to
one another since they can be balanced by no part of the
tension of the string BG.

.\ Wsin0^Fcos0,

F
.\ tan^ = -:^,

which determines 0y and the position of equilibrium is
known.

The student will remark that the tension along the string
WB' C is not of course the same ali along ita length ; for the
eflFect is the same as it would be if GB' and B' W were differ-
ent strings.

If Tbe the tension of the string at B' along B'C, T must
evidently equal the sum of the resolved parts of W and F
along B'h.

i.e. TTcos^ + i^sin^^r,
and from above TFsin ^ — jF'cos ^ = 0.

Squaring these equations and adding

which determines the tension of the string BfG.

17. The foUowing is an example of the use of the formula
giveninArt. IL

ON THE COMPOSITION AND BESOLUTION OF

ExAMPLE, A string has a weight P attached to each end,
and is then laid on a number of smooth pegs ABCD ....
Find the pressure on each peg.

The string at the point A will be kept at rest by the
tension of the string AP equal to P, and aiso by the tension

Eroduced by the weight at the other end of the string, aeting
rom ^ to JD, and also equal to P, since it will be transmitted
unaltered over ali the smooth pulìeys.

□P

The pressure on the peg A will be the resultant of these
two equal forces aeting at that point, hence if R^ be the
pressure at A and angle PAB^a^, we have by Art. 11,

JS,* = P' + P*+2PP.cosa,

= 2P' (1 + cos aj

/. jBj^ = 2P. cos-^.
Sìmilarly if JS, be the pressure on peg J?, and angle

ABC^a

J?, = 2P.cos-^;
and similarly for ali the other pressures.

F0RCE8 ACTING AT ONE POINT. 27

Sxamtnatum on Chapter IL

1. Define the term "resultant" of forces.

2. "To find the resultant of two forces actinff upon a
particle and not in the same straight line." By what name
18 this Proposition known, and how is it usually enunciated?

(N.B. In ihefollowing examples the paralUlogram of forces
ù to be aasumed as trtie.)

3. Two equal forces act npon a particle, and the angle
between their directions is 60^; find the magnitude and
direction of their resultant.

4. Two forces represented by 2 and 3 are inclined to
each other at angle 45° ; find the magnitude and direction of
their resultant.

5. At what angle must two forces 3, 4 act so that their
resultant may be 5 ?

6. A force of which the direction and magnitude is known
acts upon a particle. Besolve it into two others and deter-
mine the remaining particulars of each of the component forces
in the following cases : —

(1) When one of them is given in magnitude and
direction.

(2) When one is given in magnitude and the other in
direction.

(3) When both are given in magnitude.

7. Three forces represented by 2, 5 and 9 act on a particle.
Is it possible for them to be in equilibrium?

8. Shew that the tendency of a force in any direction is
found by multiplying the force by the cosine of the angle
between that direction and the direction of the force.

9. Two strings P4, QA, are tied to a point A and drawn
with a force 3 and 4 respectively : find what the angle PA Q
must be in order that the strain on A may be 6.

28 ON THE COMPOSITION AND RESOLUTION OP FORCES.

10. The resultant of two forces is 10 Ibs., one of them
18 equal to 8 Ibs., and the direction of the other is inclined to
the resultant at an angle of 36° ; find the other force, and the
angle between the two.

11. A and B are smooth ptdleys; a string PACBQ
passes over them. At a point C in the string a weight
of 10 Ibs. is attached; find what weights must be attached
to P and Q respectively that the angle A GB may be 120*
when the system is in equilibrium.

12. In example, Art. 16, find TTand T by the triangle of
forces, without the aid of Trigonometry.

( 29 )

CHAPTEK IIL

ON THE COMPOSITION AND RESOLUTION OP FOKCES ACTING

ON A POINT.

1. The student will now be suflSciently familiar wlth the
method of representing forces by straight lines, and reasoning
thereon, to foUow the jpr(x>/*of me j>arallelogram of forces.

The principle on which this proof is made to depend is
that laia down in Axiom 3 (Chapter i. Art. 6), viz. that **if
a force be applied at any point of a system of particles rigidly
connected together, its statical effect will be tne same if it be
removed to any other point of its direction rigidly connected
with the system on which it acts," and the converse of this as
stated in Axiom 4.

We now proceed to enunciate and prove the proposition.

ProP. If two forces acting upon a paritele in different
directions he represented in magnitude and direction hy two
straight linea drawn from the partide^ then the diagonal of
the parallelogram constructed upon these two straight lines will
represent the resultant of theforcejs in magnitude and direction,

Ist* As far as regards the direction of the resultant.

The proposition is evidently true for two equal forces p and
p as far as regards the direction of the resultant (CJhap.
II. Art. 8),

Assume ìt so far true for two unequal forces^ and m, and
also for two p and w, we shall shew it will, on this assump-
tion, be necessarily true for the twop and m + ».

30 ON THE COMPOSITION AND RESOLUTION OP

Let A be the paxticle on which the forces act. Draw AB

AD representìng the forces m and j:> respectively in magnìtude
and direction.

Produce AB^ and make BG proportional to the force n in
the same ratio as AB is to force m.

Then sìnce a force may he transferred to any poìnt of
its direction, BC will represent the force n in magnitude and
direction.

Complete the parallelogram ABED^ and draw the diago-
nal AÉ.

Then, by hypothesis, AE represents the resultant (jB sup-
pose) of ^ and m in direction.

And, hy axiom, the point of application of this resultant
may be supposed to be at J?, a pomt in its direction rigidly
connected with A.

Now since the force R when acting at A is equivalent to
the two p and m, when at E it may evidently be resolved
back agam to the same forces acting parallel to their originai
directions.

Produce DE to L, BE to E.

Then instead of the forces p and m at A,

we bave a force m acting along EL at E,

and a force p acting along EH at E,

F0BCE8 ACTINa ON A POINT. 31

We may again change the points of application of these
two forces so that

971 be snpposed to act at F

2xAp at B*

Complete the parallelogram EBCF and draw the diagonal
BF.

Then, by hypothesis, BF represents in direction the re-
sultant of the two forces^ and n acting at B.

And this resultant mar be supposed to act at jP a point
in ita direction, instead of A,

At F it may be resolved again into the forces p and n.
We have now removed the forces p and w -f n which acted
at A to F.

Therefore jPmust be a point in their resultant.

Whence the direction of the resultant of p and m + w is
along the diagonal of the parallelogram constructed on the
lines representing them.

Now, we know the proposition to be true for the direction
of the resultant of two equal forces^ and^.

Since therefore it is true for p and^, and p and /?, by the
preceding it is true iorp and^ +jp or 2p.

And since it is true for j? and j?, andj? and 2p,
it is true for^ and 2p +j? or 3p,
and so on ; therefore generally it is true for

p and rp, where r is a whole number.
And again, since it is true for rp and jp, and rp and^,

it is true for rp and 2p,

and so on by similar successive deductions it may be proved
to be true generally for

rp and «p, where « is a whole number ;

therefore it is true for ali commensurable forces, i. e, for ali
forces the ratio of whose magnitudes can be expressed by the
ratio of two whole numbers.

32 ON THE COMPOSITION AND KESOLUTION OF

This proves the proposition, as far as regards the direction
of the resoltant.

Note. That this proof extends also to ìncommensurable
forces (such as the forces V3 and 2, whose ratio cannot be re-
presented by two whole numbers) may be shewn in this
way :

Let AB, AC represent in magnitude and direction two
of the commensurable forces.

Complete the parallelogram A GDB» Draw AD.

Then we bave to shew that AD represents iu direction the
resultant.

If it do not, let any other line as AE represent that
direction.

Divide A C into a number of ecjual parts, such that each is
less than ED.

Mark off along CD parts each equal to one of these.

Then one division will evidently fall between E and i), at F
suppose.

Join AF, and draw EK, FL parallel to A C

The forces represented by AC, AL are commensurable.

Therefore, their resultant acts along AF (by Prop.),

i. e. the resultant of two forces A (7, AL lies further from
A C than the resultant of the two A C, AB^ AB being greater
than AL^ which is manifestly absurd.

FORCES ACTING ON A POINT, 33

Therefore AE is not the direction of the resultant of A C,
AB.

In the same manner it may be proved that no other
direction but AD can be that of the resultant of those incom-
mensurable forces,

Whence onr proposition is trae for ali forces as far as the
direction of the resultant is concerned.

2ndly. As regards the magnitìide of the resultant.

Let-^C', AB represent in magnitudo and direction two
forces P and Q.

Complete the parallelogram BG^

Then, by what has already been proved, the diagonal AD
will be the direction of the resultant of P and Q.

Produce AD backwards towards ZI

Then if we suppose a force equal in magnitude to the re-
sultant of P and Q to act along AK^ this force and P and Q
will be in equilibrium.

But if any number of forces be in equilibrium any one is
equal and opposite to the resultant of ali the rest.

Therefore the force P must be equal and opposite to the
resultant of the force acting along ^A'and Q.

R. S. 3

34 ON THE COMPOSITION AND EESOLUTION OF

Produce GA to JT, and make -4-Fequal to AC.

Then AF represents in magnitude as well as direction the
resultant of Qy and the force along AK.

Join FB^ and draw FG parallel to AB.

Then since FAGìb parallel to jB2>,

2,iiàFA=AC=-BD]

therefore also FB is equal and parallel to AD.

Now AF represents in magnitude and direction the re-
sultant of Q and the force acting along AK.

But the forces represented in magnitude and direction by
the lines A (?, ABy have a resultant acting along AF.

Therefore AG must represent in ma^itude the force
which we supposed to act along AK. jBut this force is
equal and opposite to the resultant of P and Q.

Therefore the magnitude of the resultant of P and Q

^AG^FB^AD.

That is, the magnitude of the resultant of P and Q is re-
presented by the diagonal of the parallelogram, of which the
sidQg represent the magnitude of the forces P and Q.

Therefore, if two forces, &c. q.e.d.

We have now established the truth of the paralleloeram
of forces upon evidence of the same kind as that on whicn the
Theorems of Euclid rest.

2. The parallelogram of forces is sometimes stated in a
form in which it is known as the triangle of forces. We pro-
ceed to enunciate it.

Observation. The student must remember that ali the
forces with which we are concemed are supposed throughout
this trenti se to act in one piane, and that piane to be coinci-
dent with the piane of the paper. For brevity, we shall in
the next and ali other investigations leave this to be under-
stood. Of course ali our operations are supposed to be carried
ou in the same piane.

POECES ACTING ON A POINT. 35

Prop. If three forces acting upon aparticle heq^ it at resi,
and if three linea he drawnparaUel respectively to the directions
of the forces, the sides of ine triangU so formed will he proporr
tional to the magnitvdes of the forces.

Let be the particle,

OP, Q, OR the directions of three forces P, Q, and B,
acting upon it and keeping it at rest.

Take Op, Oq representing P and Q respectively in mag-
nitude.

Complete the parallelogram Oprq.

Join Or, then, by the parallelogram of forces, Or is in the
same straight line with OB.

And the lines Op (or rq), Oq, Or are proportional to the
magnitudes of F, Q, B respectively.

Now draw three straight lines AB, BG, J[(7parallel to the
directions of P, Q, B respectively, that is, parailel to rq, Oq,
Or, then the triangle ABC so formed is evidently similar to
the triangle Orq ;

/. AB : BC : AG^rq \ Oq i Or

^P : Q : B.
Therefore, if three forces, &c. Q, e.d.

3. The converse of this proposition we enunciate as
foUows :

3—2

36 ON THE COMPOSITION AND EE80LUTI0N OP

Pkop. If the sidea ofa triangle taken in order rypresent in
magnitude and direction three forcea ojcting on a particle, those
ihreeforces are in equilibrium.

Observation. By taken in order we mean that ìt AC (see
preceding figure) represent a force actìng from A towards C,
the other two forces must be supposed to act from C towards
jB, and B towards A respectively ; this is expressed by reading
the sides, AC, GB, BA, not AC, AB, BG or AC, GB, AB.
In neither of which last two cases would the proposition be
true.

This proposition may be easily proved as follows :

Let be the particle, draw OP, OR, OQ parallel respec-
tively to 5J., ^0, GB,

Take Op = BA, Og = GB.

Complete the figure as before.

Then Or must be in the same straight line with OR.

And the triangles ABC, rOg are similar and equal, but
the forces represented by the sides of the triangle rOg are in
equilibrium by the parallelogram of forces,

Therefore, if the sides of a triangle, &c. q.e.d.

Problem. The foUowing is an example of the use of the
triangle of forces.

A string AOB, fastened at -4 and -B, passes through a
smooth ring to which a weight
TFis attached; shew that in the

{)OSÌtion of equilibrium the vertical
ine through will bisect the angle
AOB.

We have here three forces acting
at the point — the weight, the ten-
sion of the string along OB, and the
tension along OA.

Produce WO to any point G.
Andat (7 draw CO parallel to OA.

FOBCES ACTING ON A POINT. 37

Then the sides of the triangle GOD will, by the trianffle of
forces, be proportional to the magnitudes of the three forces
acting at ; i. e. CO to the weight, OD to the tension along
OBj DO to the tension along DC.

But the tension along OA must equal tension along OB,

(Since A OB is the same string and the tension of every
part of it is therefore the same) ;

/. OB^DC;

.-. angle 00 = angle i> 0(7,

but angle OCB = angle COA ;

.-. angle BOC=^ angle A OC.

i. e, the two Itmis of the string will be equally inclined to
the vertical.

4. The triangle of forces is extended further in a propo-
sition known as the pólygon of forces^ which we proceed
to give.

Prof. If the sides of a polygon taken in order areparallel
in direction and proportional in magnitude respectively to a
number of forces acting upon a particle; then that system of
forces will heep the partiate at rest,

lietABCBEFhe a polygon whose sides ^5, BC, CD...

represent respectively in direction and magnitude a number
of forces acting on a particle.

Join^a, ^A^^-

38 ON the; composition and kesolution op

Then the forces represented "by the linea AB, BG, GA, are
by the triangle of forces in equilibrium. Therefore the third
side AG noi taken in the same order will represent the re-
sultant of AB, AG. Hence we may remove -45, BG, and
substitute a force represented by AG.

Similarly AD represents the resultant of A G, GD ;

i.e. ofAB,BG, GD;

and similarly -4^ represents the resultant oi AD, DE;

i.e. oiAB,BG, GD, DE.

And we bave remaining three forces represented by the
sides of the triangle AEF taken in order.

And a similar proof would hold if the polygon had any
other number of sides.

Therefore, if the sides, &c. Q.E.D.

Cor. It is evident in the above proposition that any one
side taken in the opposite order to the rest will represent the
force which is the resultant of ali the rest.

5. The converse proposition, viz. If any number of forces
acting upon a paritele heep it at rest, and a series of linea he
drawn parallel in direction and proportional in magnitude to
thesefbrces respectively, the lines so dravm willform a polygon,
is easily demonstrated ; and is left for the student to work out
in question 4 of the examination at the end of this chapter.

6. By means of the parallelogram of forces we may deter-
mine the resultant of any number of forces {P^P^P^ ... sup-
pose) acting upon a point. For by taking lines to represent
these forces in magnitude and direction, we may determine
the resultant [R^ of any two of them {P^P^, and similarly
we may determine the resultant of jB^ and another of the forces
P3 and so on, until we bave reduced them ali to a single force
R, of which the direction and magnitude will be known from
the line representing it.

Or to state this in another shape ; fìrom the parallelogram
of forces we may find the Gondittons of Equilibrium of any
number of forces acting on a particle. For suppose there are

FORCES ACTING ON A POINT. 39

any n forces P^P^Pg ... P», then detennining the resultant JB
of {n — 1) of tnem, P P, ... P^- , the straight line represent-
ing R must be equa! in length and opposite in direction to
that representing the remaining force P^. In other words,
jR must be of the same magnitude as P^, and be such that
the algebraical sum of R and P^ = 0. These are the two con-
ditions of equilibrium.

The two foUowing propositions enable us, by the use of
Trigonometry, to place this extension of the parallelogram of
forces to the general problem of forces acting on a particle
in a more useM shape.

The principle upon which these propositions rest, is that
if a force (see Art. 13, Chap. li.) be resolved in two directions
at right angles to one another, neither of these resolved parts
can have any tendency to draw the particle in the direction of
the other. These resolved parts represent the whole of the
force in each of those directions respectively, and in the case
of equilibrium each of these resolved parts must be equal
to O.

7. Pkop. To find the magnitude and direction of the
resultant ofany number of forces acting upon apoint.

Let P^P^P^ ... be numerical quantities representing the
magnitudes of the forces respectively.

Draw any two straight lines -4a?, Ay at right angles to one
another.

Through A draw Ap^ parallel to the direction of one of the
forces Pj. Take Ap^ proportional to P^.

40 ON THE COMPOSITION AND BESOLUTTON OP

Let B^ be the angle the direction of P^ makes with Ax.

Then the resolved part of the force Pj in the direction Ax
is evidently = Ap^ cos 0^ = P^ cos 0^,

And the resolved part in the direction Ag^
= Ap^ sin 0^ = Pj sin 0^.

In like manner the force P,, whose direction makes an angle
0^ with Ax, is equivalent to

PjCos 0^ acting in the direction Ax^
PjSin^g Ay.

And so on for any number of forces.

Hence, adding together ali the forces acting in the sanie
direction, we fina the forces P^ P^ ... are equivalent to

Pj cos 0^ + Pj cos ^j + . . . acting in the direction Ax,
and Pj sin 0^ + P, sin ^^ + ... acting in the direction Ay.

Tor shortness' sake let

PjC08^j + P,C0S^,+ ... =X,

P, sin ^, +P, sin 0^ + ... = F.

Now let jB represent the magnitude of the resultant of the
forces P^P^... and be the angle it makes with Ax\ then

jB cos will be the resolved part of jB in the direction Ax,
jBsin^ Ay.

And since X and R cos both represent the whoU amount
of force acting in direction Ax, we have

and similarly

JScos^ = X,

It%m0^Y.
Y

Therefore tan ^ = v >

and

FORCES ACTING ON A POINT.

Two expressions which detennine the direction and mag-
nitude of È respectively,

8. PROP, To find the condttwns of eqmltbrium of any
numher offorces acting on a particle.

Suppose the forces to he ali redueed to one (-B), as in
the preceding proposition, then in the case of equilibrium
jB = 0, or

an equation which cannot be true unless

X=0 and r=0,
or Pj cos ^j + P,co8 0^ + ... = 0,

Pj sin 0^ + Pj sin ^^ + ... = ;

two conditions of equilibrium, which may be expressed
by saying that the sum of the forces resolved in any two
directtons at Tight angles to one another must each equal
nothing.

9. We now give some problems solved by the method
indicated in the two preceding articles :

Prob. I. Two forces acting upon a particle are inclined
to one another at an angle 135°; fnd the ratio between them
when the magnitude of the resultant is equal to that of the
smaìler force.

42 ON THE COMPOSITION AND BESOLUTION OF

Let P and Q be the two forces of which P is the less.
Let A be the particle, Ax the direction of the force P.

Through A draw Ay perpendicular to Ax.

Then if 6 be the angle the direction of the resultant makes
with Ax, since in this case the magnitude of the resultant
= P, resolving the forces along Ax, Ay respectively, we
have

P-f- $cosl35' = Pcos^ (1),

and Qsinl35° = Psin^ (2).

Squaring and adding (1) and (2)

P* + 2P^.co8l35'+^ = P';
/. $ + 2Pcosl35'*=0;

••• « = ^^-72'

(2_a/2
P" 1 "

Prob. II. A string A GB is fastened at its extremities
to two peqs A and P, and passes through a smooth ring C,
To this ring a weight W is attached, find the position of equi-
librium.

Suppose the figure to represent the system in a state of
equilibnum. Then the point C is kept at rest by three
forces, the weight W ana the tensions of the string AC and
CB respectively. And these tensions are equal to one another
(each equal T suppose) since A CB is one string.

Let and <f> be the angles which BC, A C make respectively
with the horizon.

Then resolvinff the forces at G along a horizontal and
vertical line, we nave

Tcos0+ Tcos<f>^0 (1),

TsìnO+TBÌn^- Tr=0 (2),

FORCES ACTING ON A POINT.

Here we have two equations contaìning three unknown
quantities 6, ^, and T, we must therefore obtain as many
more equations from the geometry of the figure as will make

^-••55-0

A^.rr::::.

9 •

the number of equations altogether as many as the number of
xmknown quantities involved.

Let AB=c, and let a be the angle AB makes with the
horizon.

Let l be the length of the string, and let CB=x. Then
from the geometry of the figure

e 008 a = a: cos ^ — (? — x) cos <f>
e sin a = aj sin ^ — {l—x) sin <{>

(3),
(4).

The two statical equations (1) and (2), together with the
two geometrical eauations (3) and (4), give us four equations
involving four untnown quantities 0, <f>, T and a?, and will
therefore suffice for the solution of the problem.

From (1) cos ^ = — cos ^ = cos (180 — ^) ;

/. ^=180-<^.
Substituting for ^ in (3) and (4),

c.cosa = a;cos^4- (? — a;) cos^,

44 ON THE COMPOSITION AND BESOLUTION OP

i.e. c. cosa = Zcos 6 (5),

and e . sin a = aj sin ^ — (Z— aj) sin 6^

i.e. c.sina = 2ajsin^ — Zsin^ (6).

From (5) we know 6, viz.

^ ^xfc. cos a\

and from (6) we know a;, viz.

— ^ f e sin g ì

~2l(P-c*cos»a)»""*"^r

Therefore the position of equilibrium is fiilly determined.

If we wish to find the strain on the string (T) we may
easily do so from equation (2).

Notice, that when a = 0, i. e. A and B are in the same

e Z . . .

horizontal line, cos d^-j and a; = - , which is evidently true.

Ohservation. The solution of the preceding problem is very
important as illustrating the method of proceeding in solving
ali statical problems, which may be thus described:

(1) Draw the figure, representing the directions of ali the
forces by arrows.

(2) Write down ali the equations of equilibrium (which
in the case of a particle will be two).

(3) Count the number of unknown quantitìes in these
equations, and then add as many more equations deduced from
the ge(ymetry of the figure as wiU make the whole number of
equations the same a^ the whole number of unknown quantities
involved.

The student is strongly advised to bear in mind these three
rules in the solution of every problem.

FOBCES ACTING ON A POINT.

Prob. III. Two forcea P and Q act at the extremitiea of
a rigid rod without weight, their dtrections heing incUned at
angtea a and /8 respecttvely to the rod/ find what force must he
applied to the beam to keep it at resi,

Let the directions P and Q be
produced until they intersect in
point 0. Then we may suppose
the point rigidly conpected with
the rod AB, by another rod with-
out weight, and this will not alter
the conaitions of the problem.

The forces P and Q may there-
fore be supposed to act at the point
(Axiom 3, Chap. i).

Let jB be the force which acting
at the point would equilibrate
P and Q, <f> the angle its direction
makes with AB.

Then resolving the forces at along and perpendicular to
a line parallel to AB, we have

jB cos ^ + Qcos^ — Pcos a =
jB sin <^ — Q sin )8 — P sin a =

(1),
(2).

The two equations (1) and (2) will suflSce to determine
a and <f>. We obtain from them

and

i? = P»+(y-2P(3cos(a + /3),
Psin a+ Qsin/S

tan^ =

Pcos a— Qco&fi'

If G be the point at which the direction of P cuts the
beam, then P may be supposed to be applied at G instead
of 0, and we have

AG^ sm{<l>-^l3)
AO sin<^ '

46 ON THE COMPOSITION AND BESOLUTION OF

AO sin a

and

AB sin (a + i8) '

.". AG= - — r-— : — r- — S . a (if a be the lenffth of the rod).

sm ^ . sin (a + p) ^ ^ ^

Hence the point G is known, since ^ has been ah-eady
determined.

And a force whose magnitude is iZ, making an angle ^
with the rod at the point O^ will be the one required.

Examination on Chapter IIL

1. Prove the parallelogram of forces as far as regards the
direction of the resultant.

2. Assuming the parallelogram of forces as far as regards
the direction of the resultant, shew that it holds trae for the
magnitude.

3. Enunciate the triangle of forces.

4. If any number of forces acting upon a particle keep it
at rest, and a series of lines be drawn parallel in direction
and proportional in magnitude to these forces respectively,
the lines so drawn will form a polygon.

5. The sides of a quadrilateral taken in order represent in
direction and magnitude four forces acting upon a particle.
Shew geometrically that the sum of the resolved parts of these
forces along any two straight lines at right angles are respec-
tively zero.

6. The magnitude of two forces acting upon a particle are
in the ratio of 2 to 3 ; find the angle between their directions
when the magnitude of the resultant is a mean proportional
between those of the forces. Ans. 125®. 41'.

7. In figure of Problem II. Art. 9, suppose ACW is
one string passing through the ring C which is tied to the
end of the other string Bu^ and supporting the weight TF, find
the strain on the peg B.

FpKCES ACTING ON A POINT. 47

8. j5 is a smooth ring and A a fixed peg in the same
horizontal line; a string WBA, supporting the weight W at
one extremity, is passed through B and tied to A. At a
point G between B and A another string is tied to the first
supporting the weight P.

Find the relation between P and W in order that the
vertical through G may bisect the line AB when the system
is in equilibrium ; supposing AG=1^ AB.

( 48 )

CHAPTER IV.

ON THE COMPOSITION AND BESOLUTION OP FORCES ACTING

ON A RIGID BODY.

1, HiTHERTO we bave confined our attention to the case of
forces acting upon a particle perfectly free to move ; we now
proceed to extend our reasoning to cases where the particle
iorms one of an assemblage of similar particles united together
by their mutuai attractions. When the number of such par-
ticles is infinite the assemblage is called a hody. Our inquiries
will be confined to that class of bodies known as rigid.

Def. a rigid hody ù one in whick the relative poaition
of the particles cannot be altered hy the action of any finite
force.

There is in fact no such thing as a rigid body in nature.
Some are more nearly rigid than others. Iron for instance is
more nearly rigid than wood, and wood than water. It may
at first a|)pear useless to raake investigations upon a hypo-
thesis which can never correspond with reality. Our method
of proceeding is, however, to base our calculations upon this
hvpothesis of perfect rigidity, and afterwards to ma\e such
allowance in our results, for the want of this rigidity, as expe-
riment shows to be necessary for the particular substance with
which we happen to be dealing. In this way our final result
thouffh not mathematically exact, is yet, in most cases, suffi-
ciently near the truth for ali practical purposes.

This method of proceeding is forced upon us by our igno-
rance of those forces or mutuai attractions which hold together
the particles or molecules of a solid body. We cannot intro-

ON 'THE COMPOSITION A'ND RESOLUTION OF FOBCES. 49

dabe those molecular forces, therefore, intò our conditions of
^equilibrium.

The principle whicli enables us to ignore the molecular
forces is that of the " superposition of forces," enunciateci in
•Axiom 5 (Chap. i. Art. 6). In a rigid body the molecular
forces are alwajs in equilibrium, we maj, therefore, inyesti-
gate the conditions of equilibrium of any system of extemal
forces acting on a body without considering the molecular
forces, knowing, that if this system of extemal forces be in
equilibrium, the whole will be in equilibrium,

A single particle is of insensible weight, but when an infinite
number are united together, forming a body, the weight of that
body becomeff, in many cases, a most important element in
our calculation. Stili the student must remember that neither
the existence of this weight, nor of the molecular forces, can
in any way affect the magnitude and direction of any extemal
force acting on the body. Thus, suppose -4 be a particle
perfectly free to move, acted upon
by a force P in direction AP. The
tendency of P is to make A move
in the direction AP. If ow suppose
^ be a particle of the body BDOj
and acted on by the same force.
The tendency of P is the same as
bcfore. Very probably, the body
will not begin to move in that di-
rection, because the action of P is
modified by the weight and shape of
the body. Stili the magnitude and
direction of the force P itself remains unaltered.

We may bere also remind the student that the principle of
transmission of force through a rigid body has been given in
Axiom 3 (Chap. i. Art. 6).

-2. The first proposition connected with this part of our
subject will be " to find the resultant of any two forces acting
in the same piane upon a rigid body." Such a pair of forces
may be parallel or not parallel, but since by Axiom 3,
any force acting upon a rigid body may be supposed to be
transferred to any point in its direction rigidly connected with
R. S. 4

50 ON THE C0MP08ITI0N AND BESOLUTION OF

the system ; and since the directions of anj two forces acting
upon a body and on the same piane, unless they are parallel,
must ìntersect, we may suppose such forces to be transferred
to the point of intersection of their directions, and find their
resultant, by the method pointed out in the preceding chapter.

We, therefore, proceed now to the case of two paraUel
forces.

This proposition is only second in importance to that of
the paralfelogram of forces. The " principle of the Lever"
deduced from it, is the property which enables us to extend
the science of Statics from the case of forces acting on a
particle to that of forces acting on a rigid body.

Prop. Tofind the magnttìide and direction of the resultant
of two parallet forces acting on a rigid body.

(1) Let P and Q be the forces acting at points A and B of
the rigid body respectively. Join AB. Suppose for a moment

P*Q

the body to be absent and the forces to act at the extremities
of a rigid rod, without weight.

At A and B let two forces each equal to 5 be sup^posed to
act in opposite directions along ABy these forces wiil mani*
festly not affect the equilibrium of the system.

FORCES ÀCTIKG OK À KIOID BODY* 51

Now 8 and F are two forces acting upon the particle -4.

Their resultant B, acting in direction AB suppose, may
therefore be obtained by the parallelogram of forces,

Similarly the resultant B' of Q and 8 acting at B may be
obtained,

Let the direction of B and B! be produced and meet at
suppose (for it is evident they cannot be parallel), and let
them be supposed to act at that point.

At C resolve them again in their originai directions.

Then we shall bave at G the four forces which before were
acting at A and By

viz. two equal and opposite forces acting parallel to AB^
each equa! to 8.

And a force P and a force Q both acting along a line CF
parallel to the originai directions of P and Q, and cutting
AB in F, suppose.

And the pair of equal forces 8 and 8 sreia equilibrium.

Therefore the remaining force acting along the line CFìa
the resultant sought.

Its magnitude is P+ ^.

Its direction will be known if the situation of the point F\r
known.

This may be determined in the foUowing manner :

JjtiAB-aj AF-x.

If a force equal to B were applied at A in the opposite
direction, this force and 8 and P would be in equilibrium, and
the sides of the triangle ACF^ taken in order, represent the
directions of three such forces, and are, therefore, proportional

to them ;

4—2

S2 ON XHB COMPOSITION AND RESOLUTION OP

P CF

8 X

(1).

CF
Similarly. | = ^ (2).

Dividing (1) by (2),

P _a — x

. « Q

" a F+Q'

Therefore the resultant of the two parallel forces P and Q
is a force {P+ Q) acting parallel to them at a point in the rod

distant pjf^ • « from A.

(2) If the forces P and Q act in opposite directions and P
be greater than Q, it may be shewn in the same way that
the magnitude of the resultant is P^ Q, and that the point
j^through which its direction passes is such that {ì{ AF=x)

. It will he noticed that in both the precedine cases the
'expression for x is independent of the angle which the direc-
tion of the forces makes with the line joining the points of
application ; i.e. Pwill remain in the same position however
the direction of the parallel forces may be varled. From
;this pròperty i^is called the cmtre of the parallel forces.

By this proposition we may find the magnitude and point
of applipation of the resultant of a system of any number of
parallel forces acting in the same direction. For we may
nnd the resultant [R^ of any two and its point of application,
then again we majr find the resultant (J?,) of jB^, ana another
of the forces aiid its point of application, and so on, until we
haye reduced them aU to a single resultant, acting at a knowH
point in the body. This point of application of the final resultant
will bave the pròperty poiuted out above, it will be the centre
of the system of forces.

FOBCES ACTING ON A RIGID'BODT. SS»

3. The weight of a bpdy may be conceived to be the
resultant of a number of parallel forcés, acting at different
points of the body ; for the weight of any portion of matter
18 the force with which it is attracted towards the Earth's
centre : and we may suppose ahy body to be divided into a
uumber of equal and very small portions, so small that the
whole weight ofone such portion may be conceived to act at
one point. Each one of these portions of matter will be
attracted towards the Earth's centre ; and since the- distance^
{rota the body to the Earth's centre is very great compared
with the dimensions of any body with wnich we shall be.
concemed, we may suppose ali these attractions to be parallel,
the resultant of this system of parallel forces is the weight
of the body ; and the point in the body at which this resul-
tant acts is called the Centre of graviti/ of the body. It is
evident, from what was said at the end of the last article,
that the centre of gravity will remain fixed, inwhatever
position the body be held. For it ià independent of the angle
which the component attractions make with any fixed. line in,
the body.

We shall, in a future chapter, eijter more at length into
the properties of the centre of gravity. The foUowing ex-
ampie may serve to illustrate what we bave already stated,
and will be usefiil in some of the examples added at the end
of this chapter.

Example. To find the centre of gravity of a uniform beam.
Let AB be the beam.

Take two small portions, P, P', equally distant from the
centre of the beam ; the weight of each of which w may be
supposed to act at the point P and F respectively.

The resultant of these two parallel forces will be one whose
magnitude ì^vx + w^ acting at a point between P and P' and
whose distance from P (see formula Art. 2) i

54 OK THE COKPOSITION ANO KESOLUTIOM OF

Or the residtant acts at tbe centre of the beam, and ìs e^ual
in magnitude to the sum of the weights of the small portions
PandP',

A similar result would be produced, if we took any other
pair of equal portions of matter equally distant from the
centre*

And it is evident that the whole beam may be divlded thus
into pairs of eqnal portions of matter, equally distant from
the centre 0, and which maj be taken as small as ever
we please.

Hence the resultant of the weights of ali the small portions
into which we divide the beam, is a force equal to their sum
(i. e, the weight of the whole body = TT, suppose) acting at
the centre of the beam.

Hence the centre of the beam is the centre of gravity, and
the whole weight may be supposed to act as a single force at
that point.

The student will be aware that ali bodies are not attracted
to the Earth with the same force in proportion to their bulk
or volume, or, as we express it, they are not ali equally heavy.
A small portion of lead is as heavy as a large bulk of wool.
The weight of the body depends upon its denaity. The pre-
ceding result would evidently not be true if the portion A
of the beam were made of wood, and OB of lead. W hen then
we say the beam is uniform, we mean to imply that it is
uniform not only in size from Ato B^ but in density also«

It may be easily deduced from the above example, that the
centre of gravity of a circle or square of uniform density, is
the geometrical centre of the figure.

The following is an example of the use of the formuhe for
the resultant of two parallel forces. *

FOKCES ACTTNG ON A KIGID BODY.

Examph. A uniform beam AB 20 ft. long is saspended
from a nail by a string which is fastened to the beam at a
distance of 2ft. from ita centre, a weight of 20lbs. is attached
to the other end to keep the beam horizontal. What is the
weight of the beam?

Let O be the centre of the beam.

Y
w

liu

Then we bave already shewn that the whole weight of the
beam ( W) msLj be supposed to act at G.

Let T be the tenslon of the string, the point in the beam
to which it is attached.

Then the beam is kept at rest by the three forces PF, T,
and 20lbs.

Therefore T is evidently the resultant of the other two,
and

but

W.OG = 20x OB, (by Prop. page 50),
0(? = 2, and 0B=8;

... Tr=?l— = 80lbs.

4. When the parallel forces are equal and act in opposite
directions, the formula in Art. 2 (2) shews us that the resul-
tant must be in magnitude and pass through a point at an
infinite distance from A, which evidently means, that such
a pair of forces can bave no single resultant.

56"; ON THE COMPOSITION AND KESOLUTION OP

Def. Two equal forces acting in parallel and opposite
directions are calied a couple*

The perpendìcular dlstance between their directions is calied
(he arni of the còuple.

The produci of oue of the forces and the arm is calied the
moment ofthe couple.

In the preceding proposition, Art. 2, we have supposed the
forces to act on a rigid rod without weight: but the same
holds for any rigid body which has a fixed point in the same
piane as the direction of the forces. For draw through the
fixed point a straight line meeting the directions of the forces
in A\ B* respectively ; then we may suppose the. force to act at
these points, and the problem will be reduced to that of two
forces acting at the extremities of a rigid rod.

5. Def. If the number of units of magnitude in a force
be multiplied by the number of units of length in the line
drawn from a fixed point perpendicular to the direction of the
force, the product is calied the moment of the force àbout that,
point.

In Art. 2, the point F through which the resultant of P
and Q passes is calied the fulcrum of the lever AB, The.
property known as the principle of the lever is this, that, in
the case of equilibrium.the moménts of the forces about F are
equal. We may easily deduce this property from. the result
already obtained in the case of two parallel forces, for, in fig.
Art. 2, draw Fp^ Fq perpendicular to direction of P and Q
respectively.

Then by similar triangles

Fp_FA
Fq" FB'

Bat we have alreadj seen that

P_FB . .

Q~ FA'

. I-Fi.
" Q~Fp'

FORCES ACTING ON A KIGID BODY. 57

or PxJp= Gx Jj,

the property required.

.This property is also true when the forces are not parallel,
and may be deduced in that case immediately from similar
triangles. (See Question 4 in the examlnation paper at the
end of this Chapter.)

6. The student will observe that the general problem
which we now want to solve is tofind the conditions of equili"
brvum of any number of forces acting in the same piane on a
rigid ì>ody.

From what has been said in Artide 3, it wlll appear that a
couple is the only case of a pair òf forces which cannot bave a
single resultant.

The student will therefore not be surprised to leam that the
" theory of couples " is connected with the solution of this
general problem.

The manner in which the theory of couples enters into the
general problem is shewn in the foUowing proposition.

Prop. Any force acting on a rtgid body ù equivalent to thè
same force acting in a parallel direction at any given point in
the body^ togeth&r with a couple whose moment is the moment of
the originai force about that given point,

Thus suppose BP to be the direction of a force P acting on
a rigid body, A a given point rigidly connected with P,

At A suppose two equal and opposite forces act in a direc-
tion parallel to P and each equa! to P. This will not affect
the system.

ON THE C0MP08ITI0N AND BESOLUTION OP

Draw AB perpendicular to the direction of the originai force.
Then we have the force P acting at A, and a couple whose
moment is PxAB, instead of the originai force P.

Hence if we have any number of forces acting on a rìgid
body in one piane, we may transfer each one to a given point,
introducing a series of couples whose moments are equal to the
respective forces about that point, and whose arms ali have
one extremity at that point.

The forces acting at one point may ali be reduced to a single
force, and the solution of our general problem will resolve
itself into finding the resultant of the series of couples which
we have introduced.

The method of doing this is indicated in the two following
propositions.

7. Prof. The effect of a couple ù not aUered hy tuming
ita arm ahout one extremity %n the piane in which it acts.

Let PP be the two forces of the couple acting at an arm AB,

Through A draw AB making any angle with AB. Make
AB = AB.

\B'

At ff apply two equal and opposite forces, P\ P" each
equal to P, and. acting at right angles to AB.

Apply two others in the same way at A, as in the figure.

FORCES ACTINa ON A RIGID BODY.

Let the directions of P ai B, and P" at B* meet at C.
Join AO.

Now the triangles ABG, AB'G are evidentljr equal in
ali respects.

Therefore AChì&ecìs the angle BCB\

And also the angle BAB**

But since P* and P acting at C are equal, their resultant B
will bisect the angle BCB\ i.e. act along the line AC.

And similarly the resultant R of P' and P at .4, equal to
the former, will act along the line A Cn

These two resultants will therefore neutralize each other,
and there will only remain the two forces P', P', acting at the
arm AB\ That is, the couple with the arm AB has been
tumed through the angle BAB'y without altering ita effect.
Therefore the effect of a couple, &c. q.e.d.

8. The preceding proposltion will enable us to prove the
foUowing.

Prop. Two couples whoae arma have a common extremity
and whoae momenta are equal have the same statical effect,

liCt the forces of one couple be each equal to P.
other Q,

Let A be the common extremitj,

/? p

Turn one couple about A until ita arm coincides with the
arm of the other. This, by the preceding proposition, will not
alter the effect of either.

6(X ON THE COMPOSrnON AND BESOLUTION OP

. Let then ABy AC bé the arms of the couples respéctively,
as in the figure.

Then since the moments of the couples are equal

P.AB^Q.AG.

Now P at -4 and Q at C, two parallel forces, are equivalent
to a force P+ Q acting in a direction parallel to that of P and
^ at a point J) such that

And, slmilarly, Pat D and Q at A, are equivalent to a
force P-^- Q acting in the same direction at a point D\ such
that

•^^"p+G'

.\AD?=-AD')

therefore the two couples are reduced to two equal and oppo-j
site forces at the point D. Thej are therefore in equilibrium.
The moments of the two couples havebeen here itkQu opposite
in direction. It foUows directly that if they be in the sàrné
direction, the statical eflfect of the two will be the same,

Therefore two couples,, &C. , Q. E. D.

It is evident from this proposition that if one extremity of
the arm be given, the statical effect of a couple depends
entirely on its moment.

The intensity of a couple is numertcally estimateci by the
product of the number of units in the force and the number of
units of length in the arm, which may be clearly shewn as
foUows,

The intensity of a couple must be measured by roeans of a
couple chosen as the unit of couple. Let this be one of which
the arm is a unit of length and the force a unit of force, such
a couple will be represented by the figure 1. If another
exactly similar couple be supei^osed on this, the effect will bé
doabled, and the same as if we had a single couple whose ^rm

T0RCE8 ACTING ON A RIOIO BODT. '61

-was unity and force 2 ; and this couple being two nnit-couple»
will be represented by the figure 2. Similarly, if n such
couples were superposed, the resultant couple wìU be one whose
arm is unity and force n, and it will be numerically estimated
by n, i.e. by the numerìcal vaine of the force.

Now let the couple to be numerically estimated be one
whose arm contains m units of length and whose force P units
of force, whose moment, therefore, is Fm.

By the preceding prpposition we may remore this couple
and substitute another whose arm is unity, and whose force is
such thatthe moment is Fm, The force of this couple is Fm,
and it has already bèen shewn that Fm represents the intensity
of such a couple, therefore Fm represents also the intensity of
the couple whose force is F and arm m, That is, a couple is
always numerìcally estùnated by the product of the force and
the length of the arm.

9. We are now in a position to solve the problem men-
tioned at the end of Art. 6, viz.

Prop. To find the resultant ofany numbèr of coujples, whose
arms have a common extremtty*

Let P, P^, Pg ... be the forces of the couples respectively,
P^Piì Pi"' *^^ cprresponding arms.

Let ali the couples be tjamed about the. common extremity
until ali the arms coincide.

Now let the couple whose moment is F^p^ be removed and
another whose arm is equal to p, and whose moment is equal
to P^Pi, be substituted for it. Let P/ be the force of this couple.

Then ^>. = i\ft;

In the same way we may substitute for Pj^^,, a couple acting
at arm p, and whose force will be

62 OK THE COMPOSITION AND RESOLUTION OP

and so on we maj remove ali the other couples until we have
got instead of them a series ali of whose arms coincide and
equal p. It is evident that ali these superimposed couples
will be equiyaleut to a single couple acting at an arm p. If
8 be the force at the end of this arm we haye

P P
or 8p=^Pp + P^p^ + Pjift + &c.

i.e. the moment of the resultant couple is equal to the algebra*-
ical sum of the moments of the compound couple,

It will be noticed that there are two ways in which a couple
may tend to tum a body, if we cali one of these ways positive,
the other must be called negative. It ìs usuai to cali these
couples positive which tend to tum a body round in the
direction in which the hands of a watch move, those that
tend to tum the body in the opposite direction must have a
negative sign prefixed to them in ali algebraical expressions
where they occur.

10. We are now able to solve the general problem :

Tofind the conditions of equilibrium ofany number ofjbrces
ctctinff upon a rigid body in one piane.

Let P, P^, P„ be the forces*

Take any point A in the body, let 5, 0^^ 6^, be the

angles which the directions of the forces make respectively
with a fixed line passing through this point.

Let o,^j, |?j, be the lengths of the perpendiculars from

A on tne directions of the forces.

Then, by Art. 6, we may remove P to -4 introducing a
couple whose moment is Pp^ and the extremity of whose arm
is at A.

And similarly, with ali the other forces.

Now let R be the resultant of ali the forces at A^
^ the angle it makes with the fixed line.

FOBCES ACTING ON A RIQID BODY, 63

Then by Art. 7, Chap. ili.,

J? sin ^ = P sin 5 + Pj sin 5, + P, sin (9, +

Also if M be the moment of the resultant couple we shall
have also by Art. 9,

Jlf=i5?+P,p, + P,;?,

In the case of equilìbrlum we must have J? cos ^, J? sin ^
each equal to 0, as in the case of the equilibrium of a particle,
and also M—^ (for since a couple cannot be counteracted
by ^ force the force and the couple must separately be zero),
Whence we have

Pcos 5 H-P^ cos ^j +P,cos^a + =0 (1),

Psin^ + P,sin(9, + P,sin(9,+ =0 (2),

i^ + P,;?, + P,^,+ =0 (3).

Three conditions of equilibrium for the rigid body.

11. The following problem will illustrate the use of these
equations of equilibrium :

Problem, A uniform beam -4P, of weight TF, hangs verti-
cally from a peg (7, by a string attached to the end A of the

->F

beam, the other end of the beam is then acted upon by a
horizontal force P, of such magnitude that the beaifi màkes

64 ON THE COMPOSITION AKD RÉSOLUTFON OP

an angle of 60** with the horizon. Find the strain on the
string CA,

Let the figure represent the beam in the position of equi-
librium. Let T be the strain or tension oi the string. ÙA,
If G be the centre of the beam, the whole weight W may be
Bupposed to act at that point (bj Art. 3).

Then the beam is kept at rest bj the three forces T, W,
and K

For our previons reasoning has shewn ns that these three

forces may be ali supposed to act at one point of the body,

'A for instance, proviaed we introduce couplea, whose forces

shall be respectively those acting on the body, and ali of whose

arms shali bave one extremity slì A,
. . •

Draw a horizon tal line through A and produce G^TFthe
direction of the weight to meet this line in m,

Draw An perpendicular to the direction of F.

Then we may conceive the beam to be kept at rest by three
-forces T, W, F, acting in their proper direct ions at A, and
two couples W acting at arm Am, and F acting at arm An.
And, besides the two equations of equilibrium of point A
obt^ined in Chap. ili. Art. 8, we bave also the third condition
that the algebraical sum of the moments of the couples about
A must be nothing. Hence if ^ be the angle CA makes with
the horizontal line through A,

TcosO-F^O (1),

Tsm0-W=O (2),

W.Am-F.An = (3).

These are the three equations of equilibrium.

Since the angle GAm = 60, if 2a be the length of the beam,
we bave

Am = a cos 60** = - ,
atfd . u4n = 2a sin 60° = a V3,

FORCES ACTINia Olf A RIGID BODY. ' 65

and equation (A) becomes,

TF. - 2 //3 . i^= (4).

The equatìons (1), (2), and (4) contain three untnown
quantities F, 0, and T, and will therefore be sufficient for the
solution of the problem without the aid of any geometrica!
equations. In this case we have to find T.

From (1) and (2) ;
From (4) ^=^^5

T' = (1 + 1)PP;
^^ 2V3 ^ ^

the negative sign of the radicai being evidently not applicable
to the case.

If it be required to find the position the beam would take
when in equillbrium we have from. equations (1) and (2)

tan^ = -j^,

whìch determines 0. And the length of the string CA being
known, the distance of A firom the vertical through C is also
known ; and the position of equilibrium determined.

12. The preceding problem has been worked with a vìew
to remind the student of the process of reasoning by which
the three equations of equilibrium for a ririd body, acted on
by forces in one piane, have been obtained. In practice we
may lose sight of this process, and proceed at once to deduce
our results from the three equations of equilibrium, afi in the
foUowing problem.

Problem. A uniform beam BCo{ weight W moves freely
about a hinge (7 in a vertical piane. A strmg JBDP is attached
to the end n of the beam, and passes over the smooth pulley

B. s. 5

ON THE COMPOSITION AND EESOLUTION OF

D supporting the weight P at the other end, G and D are in
the same horizontal line, and GB equals GD. Find the
magnitude and direction of the whole force exerted on the
hinge when the beam is in a position of equilibrium.

Let R be the whole force on the hinge, ^ the angle it
makes with the horizon, when the beam is in the position of
equilibrium, the angle BGD at the same time ;

BG^GD^2a.

Then the beam is kept at rest by the three forces ;

W acting verticallj at the centre of the beam ;
P the tension of the string BD ;
R the action at (7,

Therefore resolving horizontally and vertically, and taking
moments about (7, the three equations of equilibrium will be

JBcosó-Pcos GZ)P = (1),

jBsini^ + PsinCLDP- Tr==0 (2),

Fa cos ^ - P. 2a sin GBD = 0, or since GBD = GDB,

Facos^-P.2asin(7i?P = (3).

POECES ACTING ON A RIGID BODY. 67

Here we have four unknown quantities involved in these
three equations, .

JB, <j>, e and angle GDB;

we must therefore procure another relation between them
before we can solve tne problem ; the geometry of the figure
gives US

angle GZ?5=90--| (4),

which gives us such a relation as was required,
Substituting from (4) in (1), (2) and (3), we have

R co8^ = Psin- , (5),

^sm0= PF-Pcosl (6),

TT, 008 ^ = 2P cos ^ (7),

from (7) we may determìne 0, thus

e \ .^ e

WU cos'- - l) = 2Pco8 1 ;

^e p e i

••• C08---^C08-=-;

e p

,'. cos - = -rr-,

2 2W

*v (rfr«+2.

6
Whence, we know - , and /. d ;

dividing (6) by (5),

Tr-Pcos|
tan0 = g— ;

P.siuo

which determines ^, and substituting for^ and d in (5) or (6),
R also will be known.

5—2

68 ON THE COMPQSITION AND RESOtUTION OP

Examination on Ghapter IV,

1. Enunciate the principle of the superposition offorces;
and explain how in investigating the conditions of equilibrium
of a rigid body we are able to leave out of consideration
the forces which hold together the particles of which the body
is composed.

2. Two forces act in parallel directions at the extremities
of a rigid rod without weight, find the position and magnitude
of a third force necessary to keep the rod at rest.

Ex. Two men cany a weight between them of lOOlbs.
slung on a pole, the weight is 1 ft, nearer one man than the
other, and the pole is Gft. long. How much of the weight
does each man support ?

3. Define the Centre of Gravity.

Ex. Assuming the centre of gravity of a uniform beam to
be its geometrical centre; find the centre of gravity, Ist of
a square, 2ndly of a circle of uniform thickness and density.

4. If two forces, not parallel, act at the extremities of
a rigid rod, and be the point where the direction of the
resultant of these forces cuts the rod, shew that the monients
of the forces about are equal.

5. Define a couple. Explain the necessity of introducing
the idea of a couple into mechanical investigations.

6. Explain how the inteusity of a couple is estimated
numerically,

7. Define the moment of a force about a point.

Ex. Two men standing on opposite sides of a vertical
post pulì at it by means of ropes fastened to the top of the
post. Given the height of the post and the length of the
ropes, compare the strength of the men when the post remains
vertical.

8. A string having its extremities fixed to the ends of an
uniform rod, of weight TF, passes over four amooth pegs in the

FORCES ACTING ON A RIGID BODY. 69

same vertical piane, so as to form a regular hexagon ; the rod
which is horizontal being one of the sides ; find the tension
of the string and the vertical pressure on each peg,

9. A uniform beam PQ hangs by two strings AP, BQ
from any two fixed points A and B\ shew that when there
is equilibrium the tensions of the strings are inversely as the
sines of the angles at P and Q.

( 70 )

CHAPTER V.

ON THE COMPOSITION AND EESOLUTION OP FORCES ACTINQ
ON A SYSTEM OP TWO OR MORE EIGID BODIES.

1, When two bodies are in contact and at rest, ìt is
evident that the pressure exerted by one must be exactly
returned by the other.

Hius suppose a book ìs lying on the table, the table exerts
exactly the same pressure on the book that the book does on
the table. The contact, in fact, gives rise to two equal and
opposite forces which equilibrate one another. The same
would be trae if we were to press the book down with the
hand, or place another body upon it. It is usuai to cali the
two forces arising in this way action and reactton. We may
cali which of the two we please action, the other will then te
the reaction. The relation here pointed out between two such
forces is expressed by saying

Action and reaction are egual and opposite.

Action and reaction may arise in many other ways besides
contact. For instance, if a weight te suspended by a string
from a nail, the nail will exert a force equal and opposite to
that of the weight. We bave already, in the preceding
chapters, assumed in such cases that action and reaction are
equal and opposite.

We are at present concerned only with cases of contact.
We will consider first the contact of two surfaces, one or both
of which are smooth.

2. A smooth surface is one which cannot exert any force
by contact except in a direction perpendicular to itself. It

ON THE COMPOSITION AND RESOLUTION OP FOBCES. 71

will be seen by tbe very definltìon of a smooth surface that
when two piane surfaces, one of which is smooth, are in con-
tact, the action and reaction must act along their common
normal at the point of contact. And this will also be true
where one or both surfaces are curved. In every case the
action and reaction will be along the common normal at the
point of contact.

To make what we are saying quite plain, suppose a smooth
beam to rest on a round peg C as
in the figure. Here the beam will
press on the peg with a certain
force i2,, and the peg will press on
the beam with a force -B^, and R^
-B, will be equal and opposite.

Further, the beam being smooth / ^

can exert no rubbing force on the
peg, i. e. the action and reaction
must be perpendicular to the surface of the beam, and to the
surface of the circle at the point of contact.

Observe also, if we had to consider the equilibrium of the
Ijeam under certain circumstances, we mignt introduce the
force jB^ at right angles to the beam at the point of contact,
and then leàve the peg out of consideratlon altogether.

3. These considerations enable us to extend the investiga-
tion of the conditions of equilibrium of a ri^d body to a
system of two or more rigid bodies in contact in cases where
there is no friction between the surfaces in contact. The
foUowing examples will illustrate the method.

Próblem I.

A uniform beam AB lies in a vertical piane between two
smooth inclined planes, GB and (7-4, fina the position of
equilibrium.

« •

Let W be the weight of the beam collected at its middle
point, -Bj, JBj the reactions against the planes GB, CA re-
spectively.

72 ON THE COMPOaiTION AND BESQLUTION OP TOECES

Here we may treat the beam as though it were acted upon
hy the three forces ^^, iS, aod IF, aad diaconnected from
everj other body.

Let a, /3 be the inclinatioas of the planes to the horizon
respectively,

be the inclination of the beam to *the horizon,
2a the length of the beam.

Then, resolTÌng horizontally and vertìcally and taking
moments about B;

B^BÌna - B^ sin fi = (1),

JBjCosa + iZaC08)8- W=0 (2),

W.a.coaO- B^ 2a sin {90 - ()8 - ^} = (3) ;

from (1) and <2),

jBj cosa. sin )8 + jB, cos^ sina— Tr.sina = 0;

i. e. JB« =

TFsina

sin(a4-y9)'

from (3),

TFco8^ = 2^,cos(/9-e);
.-. Tr= 25, cos 13 + 2B^ sin )8 tan ^ ;

TT- 2.g, cos fi

25, sin /3 •

.*, tan^=s

ACTING ON A SYSTEM OP TWO OE fflOBE BIGID BODIES. 73

Substituting fer jK,,

^ sin (a + )8) — 2 sin a cos /S

tan tf = -— : : — 5 '

2 sin a sin p

BÌn (^ — a)
2 sin a sin /} '

wliich determines 0.

PeOBIìEM II.

A uniform "beam AB rests in a smooth semi-circular bowl,
the length of the heam is twice the radius of the bowl, And
the line joining the opposite lips of the bowl is horizontal.
Find the position of equilihriunu

Let G be the centre of the beam, Wthe weight of thè beam
acting at G.

The reaction at A will be perpendicular to the surface
of the bowl, that is, along the line AO, {0 being the centre
of the bowl), cali this jB^*

The reaction at C will be perpendicular to the surface of
the beam, cali this J?,.

We bave, then, to find the position of equilibrium of the
beam acted upon by the three forces B^, iZ, and W.

Let 2a be the length of the beam, a the radius of the circle,
the angle the beam makes with the horizon, x the distance
CG when the beam is in the position of equilibrium.

74 . ON THE COMPOSmON AND BESOLUTION OP FORCES

Then resolving along and perpendicular to the beam we
have, observing that angle GAO = angle AGO

JB,cos^~ TFsm^=0 (1),

^,BÌn^- Trco8^ + JB,= (2),

taking moments about A^

Tr.acos^-jB,(a; + a)=0 (3).

The Btatlcal equations (1), (2) and (3) involve the four
unknown quantities 6, a?, JB^ and R^^ we must therefore obtain
one other relation between them from the geometry of the
figure before we can. solve the problem. From the triangle
AOCwQ have

a:+a = 2acos^ (4),

which gives us such an equation as we required. .

From (3) and (4) we have, elìminating x,

TF=2i?, (5),

from (1) and (2) elimìnating iZ^, we shall get

W. sin» ^- TTcos» + B^Goa0 = 0.

Substituting for iZ, from (5), we have

cos»^ — -7 008^ = -;

.-. eoa tf = 3 {1 . ± V(33)} .

The negative sign would give a negative value of 0, and is
therefore inadmissible,

e = co8-^ri{l + V(33)}]

is the required value of 0*

ACTING ON A SYSTEM OP TWO OR MORE RIGID BODIES. 75

And from (4)

a; = a (2 cos ^ — 1)
= a[iÌH-V(33)l-]]

and the position of equilibrium is fully determined.

4. We now come to consider the case of the contact of
two rough surfaces.

In this case it is evident that the mutuai action of the
surfaces will not he, necessarilj, along the common normal at
the point of contact.

Let AB he a rough piane moveable ahout a horizontal
hinge at -4, TFa weight placed upon it.

When AB is horizontal the pressure of the piane is of
course perpendicular to the surface in contact, whether the
body be smooth or rough.

Now move AB about the hinge upwards ; if the surface
of W be smooth, the body will immediately slip down
towards -4, but if both surfaces be rough, AB may be moved
into a position AB' without the equilibrium of W being dis-
turbed.

In this case, besides the perpendicular action JB, a force /
will be brought into play, acting along the piane, and arising
from the roughness of the surfaces in contact.

76 ON THE COMPOSITION AND RESOLUTION OF FORCES

/ is the force of friction.

Suppose AB is the high^st position the piane can occnpy,
then it is evident that for eveiy position of the piane between
AB and AH^ TFwill be at rest; and a different amount of
friction will be called into existence for every position of the
piane between these limits. In every such case the friction is
an unknown force to be determined by the conditions of the
problem.

In one particular case, however, the friction is not untnown,
but has a certain relation to the normal force JS which will
suffice to determine it.

This particular case is that when the body is on the point
of motion. In this instance it is found by experiment that
the friction varies directly as the normal pressure, and is
independent of the extent of surface in contact ; that is, when
the friction is the greatest possible,

/= Mi?.

Where /^ is independent of iZ, /a is called the coefficient of
friction.

The friction being independent of the extent of surface in
contact, fi will be so also ; it will however vary for different
kinds of surfaces.

The value of the coefficient of friction for diflFerent kinds
of surfaces is determined by experiment, and the result set
down in tables.

The foUowing is an extract from tables constructed by
M. Morin :

Surfaces in contact '

Bisposition .
of flbres.

state of Surfaces.

Co-eflicient
of FrlctioB.

Oak upon Oak ....

parallel

jìdthout grease
twith grease .

•62
•44

Oak upon elm ....

parallel

without grease

•38

Gast-iroQ upon cast-iron

fiat

(fitto

•16

ACTING ON A SYSTEM OP TWO OR MORE RIGID BODIES. 77

The whole amount of action Between the two surfaces is of
course a single force, which is the resultant of the nonnal
pressure and the friction.

It will be noticed that the friction must always act in a
direction opposite to that in which motion would begin if the
bodies were smooth.

o. We now illustrate the preceding remaxks with some
examples.

Problem I.

What force acting parallel to a piane whose inclination
is a would be sufficient to keep a weig-ht W from slipping
down it ; the coefBcient of friction between the piane ana the
weight being /jl?

Here if B be the normal action against the piane, and the

body be on the point of slipping down, [jlE will be the friction
acting upwards, and if F be the required force, resoLving
along and perpendicular to the piane, we have

F+fMR" Trsina = (1),

jB- TTcosa^O (2).

Subatituting for M in (1),

jP= PTsina— Wfjb cosa
= TF (sin a — /A cosa),
the required vaine of F.

78 ON THE COMPOSITION AND EESOLUTION OF FORCES

The student wìU observe that if the body were on the point
of moving up the piane the frictiou would act in the opposite
direction, and if F* were the vaine of the force in this case,

jP'= TT (sin a + /a cosa),

a result which we obtain from the expression for F by simply
changing the sign of /li. Any force between -Fand F* wonld
keep the body at rest.

Problem II.

P and Q are two weights connected by an inextensible
string and resting upon a rough semicircle. PQ subtends
a right angle at the centre of the circle ; find the lowest position
which P can occupy consistently with equilibrium,

Let AQPBhQ the semicircle, its centre.

Let be the angle POE when P is in the stated position.

Then P is kept at rest by four forces,"viz.

The pressure R acting along the normal at P.

The tension T of the string and the friction fiR^ both acting
along the tangent at P.

And the weight of P acting vertically downwards. Eesolv-
ing these forcea along the. normal and tangent at P re-
spectively,

Psin^-5?=:0 (1),

Pcos^-/aS- r=0 : (2).

AGTING ON A SYSTEM OF TWO OK MORE BIGID BODIES. 79

Substituting for B bora (1) in (2),

. Pco8^-/iPsind-r=0 (3).

Similarly, we shall bave from the two equatìons of equi-
librìum for Q,

QBÌue + fiQcose-T^O

(4).

Eliminating 2^betwQen (3) and (4),

(P - /i4 G) cos ^ - ( C + /aP) sin e

... tan^ = ^t
G + /aP

which gives the required position.

= 0;

Pboblem ni.

A uniform beam rests with ita lower extremity on a
horizontal and its upper against a vertical piane, if /a, /a' be
the coefficients of friction between the beam and the horizontal
and vertical piane respectively, find the position of equilibrium,
when the beam is on the point of slippmg.

Let OAj OB be the horizontal and vertical walls.

B the reaction at A^
R P,

W the weight of the beam acting at the middle point G.

80 ON THE COMPOSITION AND RESOLUTION OF PORCES

the angle BA when the beam is in the required position
of equilibrium ; 2a the length of the beam.

Then the beam, when at the point of slippìng, is kept at
rest by the five fi)rcea J8, /aS, R\ ^lE and W as shewn in
the figure.

Resolving borizontaliy and vertically and faking moments
about A,

liR^B'^0 (1),

Rjffi'E- )r=0 .^ (2),

W. a CCS ^ - E. 2a sin 9-fiE 2a cq&0 = O (3).

The equations (1), (2), (3) involve only the three unknown
quantities -B, R' and 0j and will therefore suflSce to de-
termine 0j

from (1) and (2),

I + //'/^

Substituting this vaine of E in (3),

Tr.co8g-2(sing-f/^'cosg) , ^ , . 17=0;

.'. H-/A/A'-2/Lfctan^-2^^' = 0;

.'. tangrrr ^""^^ ,

2/L6

If there be no friction fifju each equal 0, and tan ^ = j: >

i.e. ^ = 0, the beam will only rest in a horizontal position
which is manifesti^- true.

The student is recommended to work now as many as
possible of the examples which will be found in the conclud-
ing Chapter of this book.

ACTING ON A SYSTEM OF TWO OK MOBE BIGID BODIES. 81

Eocammation on Chapter V»

1. Define the terms action and reaction,

2. Define perfect smoothness. Is such a thing as a perfectly
smooth body known in practìce?

3. A sphere rests on two inclined planes, shew that ìf a
and 13 be tne inclicatìon of the planes respectively ; R^^ B^ the
pressnres ; W the weight of the sphere ; then

^ Tr.sin/3 ^ TT.sina

sm(a + )8)' » 8Ìn(a + )8)'

4. Define Coeffident of friction.

5. A given body is placed npon a horizóntal table which
is moveable abont a horizóntal hinge, when the table has been
elevated to make an angle a with the horizon the body begins
to slip. Shew that the coefficient of fiiction between the body
and tne table == tan a.

6. A body is pressed against a horizóntal piane in such
a way that the whole pressure of the piane against the body
is a force 8^ making an angle <^ with the horizon. How much
of this is dueto friction?

B.S.

( 82 )

CHAPTER VI.

ON THE CENTRE OF GRAVITY AND EQUILIBRIUM.

1. In Art. 3, Chap. iv. the term centre of gravity has
been explained. It will be sufficiently evident from that
explanation that eveiy body must have a centre of gravity,
and can have only one snch centre; for a system of par-
allel forces must haye one and no more than one resultant. It
is also shewn in the same article that if the centre of gravity
of a body be fixed in any way, the body, whatever its position,
will have no tendency to tum about that fixed point.

We now proceed to discuss some other properties connected
with this suDJect.

2. If we can by inspection determine a point aronnd
which the material of a body is symmetrically situated, that
point will be the centre of gravity of the body. It has been
shewn in Art. 3, Chap. IV. that the centre of a nniform beam
is its centre of gravity. This result might have been arrived
at by the general consideration that aronnd the centre of the
beam, the material being symmetrically situated will be
equally drawn towards the earth's centre ; and, therefore, the
centre of the beam must be the centre of the system of forces
produced by the attraction of the different parts of the beam
towards the earth.

In the same way the centre of gravity of a sphere is evi-
dently its geometrical centre, for any change in the position
of the sphere can produce no chance in the disposition of the
material about its centre. Hence if the centre were fixed, the
sphere could have no tendency to tum about that point, from
one position into any other. »

ON THE CENTRE OF GRAVITY AND EQUILIBRIUM. 83

The cantre of ^avity of many geometrica! fignres may be
determined in this way. Thus tne centre of gravity of a
sanare, or a parallelogram, is the intersection of ita diagonals.
The centre of gravity of an ellipse is the intersection of its
axes major and minor.

3. We may sometimes readily find the centre of gravity
of a figure by the foUowing property.

Prof. A body is divided into two parts and the centre
of gravity of each part, and the weight of the whole body
and one part is known : find the centre of gravity of the
whole body.

Let ADG be the body,

W. the weight, O^ the centre of gravity of the portion
ABD,

W^ the weight, G^ the centre of gravity of the portion
ABC.

Here then we bave two parallel forces acting at G^^ (?«
respectively.

And their resultant will be a force W^+ W^ acting at a
point G in the line G^,G^ such that

(Art. 2, (1), Chap. iv.), which determines G the centre of
gravity of the whole body.

6—2

84 ON THE CENTRB OP GRAVITY AND EQUILIBEIUM.

'GOE. 1. If any two of the poìnts Q^y G, O^ are given, it is
evident that the tnird maj be obtained.

Cor. 2. If the body were dìvided into n parts and Gy
G^y (?g, ... (?n were the centres of gravity of the whole body,
and each of the parts respectively , it is evident that n of the
points Gr, G^y G^, ..^ G^ being given, the remaining one might
be determined.

4. It is sometimes required to find the centre df gravity of

a system of bodies.
^^ •
PiROP. To find the centre of gravity of suoli a system.

Let A and B be two bodies,

W^ and W^ their weight,

G^ and G^ their centres of gravity.

Then if we suppose G^ , <?j to be connected by a rigid rod,
without weight, or in some other imaginary manner, the
centre of gravity wiU be found as in the preceoing proposition,
and will be a point G* in the line G^^, à^ such that

We mijght suppose A and B removed and a single bodjr
whose weight was H^+ H^, and centre of gravity & substi-
tuted in their place*

ON THE CENTRE OF GRAVITT AND EQUILIBRIirgr. 83

If (7 be a third body whose weìght is W^ and centre of
gravily 6^3,

The centre of grayity of the sj^stem of bodies A, B, and C
will be a point Q in the line joining G' and G such that

W
G'G^ 3 ■ ffQ

And Bimilarly we might find the centre of grairity of a
system of any nomber of separate bodies.

Cor. 1. We might suppose the bodies A, B, (7, to be dif-
fbrent pórtions of the material of the saìne hàdj. The pre-
ceding proposition would then, with a slight modification of
reasoning, establish the fact that every system of material
particles, however arranged, must bave a centre of gravity.

Cor. 2. It is also evident from thìs proposition that the
centre of gravity of a body is not necessarily a point within
that body.

The centre of gravity of a circle, for instance, is the centre
of the circle ; the centre of gravity of a hoUow sphere is the
centre òf the sphere.

5. To find the centre of gravùy of a triangle,

Let ABGhe^ the triangle, (which is supposed to be of small
and uniform thickness).:

We may conceive this triangle to be made np of a nnmber
of rodsj'all parallel to 5(7. Let le be one of the series of rods,
the centre of gravity of he will be e? ita middle point (p. 53) ;
and d will be a point on the line jlJ? drawn.from -4 Jo the

86 ON THE CENTBE OF QRAYITY AND EQUILIBRIUM.

bisection oiBG. And the oentre of gravitar of eveiy other one
of this serìea of rods must also lìe on this line AD. Therefore
the centre of gravitj of the whole triangle must he somewhere
in the line AD.

Now we might haye sapposed the triangle to bave been
made up in the same waj of a series of rods ali parallel to the
side A Cf and then we should haye obtained the result that
the centre of ^avity of the whole triangle inust be some-
where in the line JBJE^ drawn from £ to the bisection of the
QÌàeAC.

Since, therefore, the centre of gravity of the triangle is
in the line ADj and also in the line ÈE, it must be the
point G where those two lines intersect.

We may find the vaine of QD thus :

Join DE.
Then DE is parallel to BA, (Euc. vi. 2) ;

therefore the triangle A OB^ EQD are similar ;

hence

Aa : GD^AB: ED

= 2:1,
or AQ^^.aD.

Therefore GD^\ AD.

Whence we bave an easy mie for finding the centre of
gravity of a triangle. From *an angle A draw a line to the
bùectian D of the opposite side. Tàlee DG e^tial one third
of the whole line DA. G is the centre of gravity.

* 6. To find the centre of gravity of a pyramid having a
triangular base.

Let OABG be the pyramid, ABG being the triangular
base. Bisect the edge ACìnD^ join OD^ DB.

ON THE CESURE OF QBÀTITT AND EQUILIBBIUH. 87

Take DE = \dO,

DF^ \ DB.

Then E and F are evidently the centres of gravity of the
triangulax faces AOG/A GBy respectively . Join OF, BE.

Sìnce these two lines are in one piane, they will intersect
in some point G. G is the centre of gravity of the pyra-
mid.

For we may suppose the pyramid to be made of a number
of triangles of small and nniform thickness, each parallel
to the base ABC; let aòc be one of these triangles.

Then, becanse the piane AOG meets the parallel planes
ABG, oJc, in the lines A G, oc, these lines are parallel.

And similarly AB^ aby are parallel, and also BG^ he,

Also if db be the line in which the piane ODB meets
the piane ahc^ db is parallel to DB.

Let also OF meet db in f.

Then, because AG \a parallel to (ic]

.% AD : ad^ OD : Od

^DGxdc,

88 ON THE CEKTRE OP GRAVITY AND EQUILIBRIUM.

and AD = DO; /• ad==dc.

And similarly, since

DF=IbF;

Therefore Id is drawn from the angle h to the bisectìon
of the opposite side oc of the triangle ohe, and / is a point

in hd such that df^ - db ; therefore/ is the centre of gravity

of the triangle ale.

And similarly it may be shewn that every other one
of this system of triangles of which we suppose the pyramid
composed will have its centre of gravity in the Ime OF.
And, therefore, the centre of gravity of the whole pyramid
is somewhere in that line.

And, again^ if we suppose the jpyramid to be made up
of triangles ali parallel to the face ÒAG, it may be similarly
shewn, that the centre of gravity of the pyramid must be
somewhere in the line BE.

And hence the centre of gravity of the pyramid must
be the point G, the ìntersection of the two lines BJE and OF.

We may find GF in the foUowing manner: join FF.
Since FF cuts OD, DB, in the same proportion, it must be
parallel to OB, and the triangles GB, FGFaxe similar ;

.-. OG: GF^OBiEF,

^DBiDF,
= 3:1;

.-. GF=r\oF.

Hence, we have thè following- practiòal rule for finding
the centre of gravity of such a pyramid :

ON THE CENTBE OP GRAVITT AND EQUILIBKIUM. 89

Draw a line front one of the angUs of the pyramtd to the
centre of gravity of the opposite face, a potnt in this line
distant from the face one-Jourth of the whole line will he the
centre of gravity.

Cor. 1. If the base of the pyramid be any polygon, it
may easily be shewn that the centre of gravity of such a
pyramid will be situated on the line joining the apex of the
pyramid with the centre of gravity of the base, and will
be distant from the apex f ths of the length of this line.

For if be the apex of such a pyramid, G^ the centre
of gravity of the base, join 00^. Then we may suppose
the pyramid to be made up of a number of laminae, ali
paraUel to the base, and it can be easily shewn that the
centre of gravity of each of these laminae, and therefore the
centre of gravity of the whole pyramid, is situated on the
line OG^.

Again, if we join G. and ali the angles of the base, the
pyramid will be divided into as many triangular pyramids
as there are sides to the base. In OG^ take a point G,

such that G^G ^jOG^. Through G draw a piane parallel

to the base of the pyramid. It will be easily seen that the
centre of gravity of each of the triangular pyamids, and
therefore the centre of gravity of the whole pyramid, must be
in this piane.

But the centre of gravity of the whole pyramid has been
already shewn to be in the line OG^.

Therefore, it must be the point G, the only point whìch
is both in the piane and line.

Hence we may find the centre of gravity of any pyramid
on a polygonal base in the manner stated above.

Cor. 2. The same rule may also be extended to the
case of a cone whose base is any closed curve (a circle or
ellipse for instance), for such a figure may be considered

90 ON THE CENTBE OP GRAVITY AND EQUILIBRIUM.

to be a pyramid whose base is a polygon of an infinite
nmnber orsides.

7. To find the centre of gravity of the mrface of a cone
on a circular base.

Such a surface maj be supposed to be made np of an
infinite number of isosceles triangles, each one of which
hafl its vertex in the apex of the cone, and its baae infi-
nitelj small, and coincident with the corres{K)nding are of the
base.

The centre of gravily of ea<ìh of these triangles will be
situated on the line drawn from the vertex of the triangle
perpendicular to the base, and distant \ of thewhole of
that line from the base.

Hence the whole weight of the surface may be supposed
to be coUected in a ring, running round the surface at a
distance of Jrd of the way up the side of the cone.

^ The centre of gravity of such a rinff is its centre, which
is evidently situated m the axis of the cone, and distant
from the base §rd of length of the axis.

8. The general formulae for ascertaining the centre of
gravity of any piane or solid figure, require a knowledge of
the Integrai Calculus, and, therefore, cannot be given nere.
Many centres of gravity may however be obtained by artifices
similar to those used in the preceding articles.

It is often possible practically to obtain the centre of gravity
of a piane surface very easily with a rule and compasses. For
instance, suppose it be required to obtain the centre of gravity
of the trapezium ABGD^ join A C, bisect it in 0.

ON THE CENTRE OF QfiAYiTY AND EQUILIBRIUM. 91

Join BOy DO.

Take OG^^- OB. O^ is the centre of gravity of the tri-
angle ABG.

Tgke 0(?,= - OD. O^ is the centre of gravitj of the tri-
angle A CD.

Therefore the centre of graviiy of the trapezinm is in the
line.(?j(T,.

. Similarly, the centres of gravity G^/, G^ of the triangles
ABD, BDG may be found, and the centre of gravity of the
trapezium will be in the line G^G^.

Therefore the centre of gravity of the trapeziam is the
point of intersection of the two lines G^G^ and Gì G^.

The centre of gravity of any body bounded by two parallel
planes may be experimentallv determined by snspending it
first from one point and then irom another in the way pointed
out in Art. 12 of this chapter.

We now proceed to ^ve to the student two other illustra-
tions of methods by which a centre of gravity may be some-
times found.

In ali problems where the centre of gravitv to be found is
that of a piane surface, the student may verity the result by
cutting out a piece of cardboard of the shape of the ^ven
body, determimng the centre of gravity by the method pomted
out, and then ascertaining whether the cardboard will balance
upon a needle on the centre so determined.

9. A GB is an isosceles triangle from which the upper part
has been cut off by a line db bisecting the two equal sides.
Find the centre of .gravity of the figure AahB.

This may be readily done by the property pointed out in
Artide 3.

Let G^j H^ be the centre of gravity and the weight respec-
tively of the whole triangle ACB.

G^y TFg be the centre of gravity and the weight respec-
tively of the triangle aCb.

92 ON THE CBNTRE OF GfiATITY AND EQUILIBRIUM.

Let G be the centine of gravity of the portion aABb.

G^ , G^ wìU both be in the line CD, tìrawn from G to
the Hsection of AB,

And, therefore, G will also be in that line.

Two parallel forces ( W^ — W^ and TT, act at the points
(7, 6^5, respectively, therefore their resultant W^ acts at a point
G^, such that

G G=^-

GG..

And TFT, TTj are proportional to the areas of the triangles
ABC, aòC;

/. ^2 = 7 ^1 ; also, it 18 evident that G^G^^l CD.

/. G,G^\.\.CD^\CD.

3 3

.% DG^^CD,

which determines the centre of gravity of AabB,

10. Three uniform heavy beams, such that their weight is
proportional to their leneth, are ioined toeether at their ends
lo L to fonn a triangS; find L centre%f gnivity of the
system.

ON THE CENTEE OF GRAVITY AND EQUILIBRIUM. &3

Let ABG be the triangle so formed.

Bisect the three sides in a, i, and e respectively.

Then a, 5, and e will be the centres of gravity of the tìiree
beams respectively.

Hence the problem is the same as that of finding the centre
of gravity of three heavy particles at a, i, and e equal to the
weights of the beams BUy GA, and AB respectively.

Join aCf ahy he.

Then, since he euts AB and AG in the same proportion, the
iwo triangles A GB, Acb are similar ;

/. BG : hc = AB : Ac
= 2 : 1;

i.e. he is half J5(7, and similarly ah, <zc are each haliAB,AG.

Hence, the weights at a, J, and e being proportional to the
length of the sides BG, A C, and AB, are also proportional to
he, ac, and ah respectively.

Now from h draw hV bisecting the angle cha»

Then because hh' bisects the angle eòa,

eh' : Va =^ eh : ha

= wt. of BG : wt. of BA.

Therefore V is the centre of gravity of BG and BA*

Hence we may remove the two beama and substitute a
single weight at b\ and the centre of gravity of the weight at

94 ON THE CENTEE OF GRAVITY AND EQUILIBEIUM.

V and the weight ot AG collected at 6, will be the centre of
gravity of the whole three beams. But thia centre of gravity
must evidently be in the Une hV.

Simìlarlj, if we draw ce' bìsectìng the angle hca^ the centre
of gravity of the three beams is in the line Gc\ Therefore
its actaal position is at the intersection of these two lines
hb' and ce.

11. The centre of gravity being known we can in very
many cases determine by inspection whether a body placed
upon a snrface will remain at rest.

If a body have one point in contact with a horizontal piane,
it will remain at rest if the vertical line through the centre of
gravity passes through this point of contact.

A sphere or circle, for instance, will remain at rest if placed
upon a horizontal piane. For we may suppose the whole
weight of such sphere or circle to be collected at its centre, the
vertical line through this centre will represent the direction in
which the weight acts, and will pass through the point of con-
tact with the piane ; and will there be exactly balanced by the
reaction of the piane.

Again, if a body be placed so that it have two points in
contact with the horizontal piane, it will remain at rest if the
vertical through the centre of gravity passes through the line
joining the two points of contact.

Thus, in the figure, the vertical line through (?, the centre
of gravity, passes through the line oJ joining the two points
of contact. The reactions R^ , R^ at these two points will.

ON THE CENTBE OF aRAYITT AKD EQUILIBBIUM. 95

from the nature of reaction, together exactlj equilibrate the
weight of the body. And it is evident that the bodj has no
tendencj to tnm over. It will, therefore, remain at rest.

And generally, if a body bave any number of points in con-
tact with a horìzontal piane, it will remain at rest if the verti-
cai through the centre of gravity cuts the piane at any point
within the line bonnding ali the points òf contact.

Thus, in the figure, the body has an infinite nnmber of
points, represented by the portion of its sorface ai, in contact
with the piane. It will remain at rest, since the vertical
through the centre of gravity passes through the surface of
contact. •

The figure given below represents the efièct when the body
is of such a shape and so placed that the vertical through the
centre of gravity does not pass through the surface in contact
with the piane. The whole pressure in this case will be

thrown on the point J. And the weight acting at the centre
of gravity will tum the body round this point.

. These considerations explain the reason why it is so diflScult
for a person to stand upon one leg. A person so standing
must balance his body in such a way as to place bis centre of

96 ON THE CENTEE OF GBAVITY AND EQUILIBBIUM.

gravily vertically over the sole of his foot. When, however,
he stands on both feet his centre of gravity may he anywhere,
provided the vertical through it passes through the space
mcluded between the two feet. If the feet be separated by a
considerable interval, the body may be placea in a great
variety of postures without disturbing the person's equilibrium,
or making it necessary for him to move his feet.

It will also be seen that a wall need not necessarily be ver-
tical in order to stand. The walls of lofty buildings are, in
fact, very seldom accurately so.

li AB CD be the section of a wall, (supposing the material
to be sufficiently strongly put together to be considered as one
mass), the wall may lean over until the vertical through the
centre of gravity passes through the point A. If the section
of the Wall be a rectangle, its limiting position of equilibrium
will be such that the point D may be the whole thickness of
the wall distant from the position it would occupy if the wall
were vertical.

The leaningtowers of Bologna and Pisa, in Italy, are ex-
apaples of this. The former is 134 feet high and 9| feet out
of the perpendicular, and the latter 315 feet high and 12|- feet
out of the perpendicular. They do not fall, because the ver-
tical through their centre of gravity falla within their base.

ON THE CENTRE OF GRAVITT AND EQUILIBRIUM. 97

12. If a heavy body be suspended from a point it will rest
with its centre of gravity in the vertical passmg through the
point of suspension.

Fot since the weight of the body
acting vertically at the centre of gra-
vity, and the tension of the string by
which the body is suspended, are the
only forces acting on the body, the
string must evidently lie in the vertical
through the centre of gravity, or these
forces would not equilibrate one another.

This suggèsts a method by which the
centre of gravity of a body may, in
many cases, be experimentally deter-
mined.

Let one^ end of a string be fastened
to the point A of the body and the
other to the nail 0. Let a weight W
fastened to a fine string be suspended
also from the nail by the side of the
body. Mark on the surface of the body
the direction of the line W. The line
so marked out is the vertical through 0, And the centre of
gravity of the body is some point in its direction. Now let
the string be detached from A and fastened to another point
of the bodv B, and let the vertical through be traced as
before on the surface of the body, Then the point of inter-
section (G) of the two lines so traced will be the centre of
gravity of the body.

13. Def. If a body under the action of any force be in a
position of equilibrium and a very small displacement be given
to it, if it then tend to return to the originai position of equi-
librium, that position is called one oi stable equiltbrium.

If, however, the bodv tend to move farther from its originai
position, that position is called one of unstabU equilibriwn. .

If it remain in the new position which the displacement has
given it, the position is said to be neutral.

R.S. 7

98 ON THE CENTRE OF GRAVITY AND EQUILIBRIUM.

The foUowing are examples of these three kinds of equl-
lìbrium :

A weight suspended "by a string is an example of stable
equilibrium, for if puUed slightly out of the vertical it will
tend to return to its originai position.

A penny balanced on its edge, or a stick balanced on the
finger is a case of unstable equilibrium.

A sphere resting on a horizontal table will remain in its
new position if slightly disturbed, and is tberefore in neutral
equihbrium.

Examinatian on Chapter VL

1. Define the Centre of Qravity of a body ; and find the
centre of gravity of a uniform rigid rod.

2. Find the centre of gravity of a triangle ; and deduce
that of the surface of a right cone.

3. Assuming the position of the centre of gravity of a tri-
angle, shew that that of three equal weights rigidly connected
so as to form a triangle. is coincident with the centre of gra-
vity of the triangle so formed.

4. A body is divided into two parts, the centre of gravity
of each of which is known, shew how to find the centre of
gravity of the whole body.

Two isosceles triangles are described upon the same base ;
find the centre of gravity of that portion of the area of the
greater which is not included by the less.

5. Four weights 1, 3, 5, 7, are rigidly connected together
so as to form a sanare whose side is 20 inches. Find the
centre of gravity of the system,

6. Explain under what circumstances a body placed on a
horizontal piane will remain at rest.

What will be the corresponding condition of equilibrium
when the body is placed on an inclined piane sufficiently
rough to prevent slipping?

ON THE CENTRE OP g'rAVITY AND EQUILIBEIUM. 99

7. A trian^olax board rests in equilibrium with its base
on a horizontal piane sufficientlj rougn to prevent ali sliding.
A force acts at its vertex parallel to the base. Find the
greatest valuè of this force consistently with equilibrium be-
ing preserved.

8. A cube rests upon an inclined piane rough enough to
prevent slipping. Find the incliuation of the piane that the
body may just roU down.

9. - th part of a triangle is cut off by a line parallel to the

base ; find the centre of graviiy of the remaining equilateral
portion.

10. Explain how the centre of gravity of a body may be
experimentally determined by suspending it from a point.

11. Define the terms atabUy unstabUy and neutral equi-
Ubriwm.

12. Upon a given base and with a given vertical angle
which is tne most stable triangle that can be described ?

13. A Wall whose centre of gravity coincides with that of
its vertical section ABCJD, is acted upon by a force J^'in the
piane of that section. Determine the greatest value of jPcon-
sistent with equilibrium.

Shew from the result what must be the weight of the wall
when equilibrium is just preserved in the following case : the
direction of F meets the top AS of the section at a point n

such that ^71 = ^ ft, and FnA = 45^ F= 2000 Ibs. The

height AD of the wall = 10 ft. and the width AB=4t ft.

14. A circular table rests upon a pedestal which diverges
into 3 feet. The extremities of the feet touch the floor at
points vertically beneath the circumference of the table.
Shew that it will require a weight equal to that of the table,
placed on it to upset it.

7-Z

( 100 •)

CHAPTER VII.

ON MACHINES.

1. Any contrivance whlch enables us to change the point
of application, direction, or intensity of a force may be called
a Machine,

Eods for pushing, cords for puUing, surfaces and such like
simple contrivances for changing the point of application or
direction of a force, would, according to this definition, be
machines. They are not, however, usually so reckoned.

The foUowing six machines are for convenience regarded as
simple or elementary ones :

1. The Lever.

2. The Wheel and Axle.

3. The PuUey.

4. The Inclined Piane.

5. The Screw.

6. TheWedge.

These are called the six Mechanical Pow&rs.

Some writers speak of Toothed Wheels as a seventh me-
chanical power.

Every machine, hòwever complicated, may be shewn to
be composed of combinations of these so-caUed elementary
powers.

The mechanical powers are considered, as we said, for con-
venience, simple or elementary machines. They are not really

ON MACHINES. 101

SO. It would be more correct to start with twO, the lever and
the inclined piane, as the only elementarj machines, and to
shew that the other four are only varieties of these ; but this,
though more correct, would be less conyenient.

The machines with which we shall have to do are for the
most part those which enable us to overcome a great resistance,
or lift a* great weight, by the application of a small force.

The force applled to a machine to set it in motion is called
the Power (P), and the resistance to be overcome is termed
the Weight {W).,

Most i^achines are useless except when in motion. It does
not fall within the province of our subject to consider them in
that state. But we may determine the value of the power
which would suffice to equilibrate the weight, and any in-
crease in this value of P will of course enable us to work the
machine.

The efficiency or working power of a machine will be mea-

W
sured by the fraction -^- .

It will be the principal object of the present chapter to find

W .
the value of -p in each of the mechanical powers when in a

state of equilibrium.

-p is called the Modulus of a machine.

When W\& greater than P the machine is said to work at a
mechanical advantage^ and when W is less than P at a mechani-
cai diaadvantage.

The Lever.

2. Def. a Simple or Mathemaiical Lever is a rod with-
out weight capable of moving about a fixed point.

102

ON MACHINES.

When the rod is not straight it is termed a Bent Lever,
The fixed pomt is called the Fulcnun. The power and the
wetffht are applied at two other points in the rod.

The lever is divided into three classes or orders determìned
by the relative position of these three points.

In the three foUowing figures let C be the fixed point about
which the lever tums, or ralcrum ; A the point of application
of the power (P) ; B the point of application of the weight
(W).

(l) In a lever oi first order the falcmm [G) is between
the points of application of the power and weight {A) and {B).

yì

(2) In a lever of the second order the fulcrum is at one
end, the power being at the opposite end, and the weight in
the middle.

A''

C
1\

(3) In a lever of the third order the fulcrum is at one end
and the weight at the other, the power being in the middle.

ON MACHINES.

103

■^

3. As an example of a lever of the first order we may
mention the crowhar^ used as in the figure, where TF is a mass
of rock to be raised, C a smaller piece of rock or a stone on
which the lever is placed as a fulcmm, P a hand applying the

Ì)ower. We know by experience that the efficiency of such a
ever is increased by increasing the distance between the ful-
crum and the power or by lessening the distance between the
fìilcram and the weight. We shall see presently that this is
generally true, and that any weight however great, might be
raised (at least theoretically) by any power however small, by
placing G sufficiently near to W and far from P.

Scissors zxidi Carpenters* jpmcers are examples of paìrs of
levers of the first order. The power being applied by the
hand, the common fulcrum being the joint, and the substance
to be cut or pinched corresponding to the weight. A common
pump handle; a poker between the bars of a grate raising
the coals ; are further examples of the same order of lever.

As an example of a lever of the second order we may men-
tion the crowhary when so used that its extremity rests on the

104 ON MACHINES.

ground at a point beyond that in contact witli the weiglit to
be raised ; as in the figure.

A loaded wheelbarrow when lifted before being put in mo-
tion is another example. The load is the weight, tue fulcrum
ìa at the wheel, and the power is applied by the man lifting.

The oar of a boat is also a lever of the second order. The
water affords an imperfect fulcrum, the resistance of the row-
lock corresponds to the weight, and the power is applied at the
other extremity by band.

Nutcrackers are a pair of levers of the same kind.

As an example of a lever of the third order ^ we have the
instance of a man in the act of raising a long ladder from the
ground.

Linibs of anùnals are also levers of this order. Take for
instance the human arm, and suppose the band to be sup-
porting a weight. Here the elbow-joint is the fulcrum, and
the power is applied 'by means of the muscles at a point
between the joint and the weight.

4. To fina the MODULUS of the Lever.

In the figures Art. 2, the directions of P and W are drawn
as though they were parallel and perpendicular to the rod.
This need not be the case.

Every lever, of whatever order, and whatever the directions

ON ]^ACHINES. 105

in which the power and weight act upon it, will be an instance
of a rigid body kept at rest by three forces, viz. the power,
the weight, and the reaction of the fulcrum.

Hence, by Art. 10, Chap. iv, the conditions of equilibrium
will be

(1) That the forces transferred parallel to their respective
directions to some point in the body, as Gthe fulcrum, must
be in equilibrium.

(2) That the sum of the moments of thè forces about the
same point G must be zero.

But since (7 is a fixed point it is evident that the first con-
dition may be dispensed with in this case.

And hence
if p be the length of perpendic" from G on the direction of P,

..w TT,

Pxp = Wx w,

is the one necessary condition of equilibrium.

. , TF_ perpendicular from C on direction of P
P perpendicular from G on direction of W

is the expression for the modulus or mechanical advantage of
the lever.

CoK. If the directions of P and W be parallel, as in the

figure, and we draw Cm Cw perjjendlcular respectively to
these directions, we shalt bave, by similar triangles,

Cw" GB'

106 ON MACHINES.

W CA

Therefore

GB'

A property wliich ìs expressed by saying, the power and the
weignt are inversely proportional to the lengths of the arma at
which they a^t.

We might have obtained the same property immediately
from Art. 2, Chap. IV.

5. When the forces are not parallel we may obtain the
expressìon for -p independently of the general eq'uations of
equilibrium for a rìgid body.

For let AP, BWhe the dlrections of the power and welght
respectively, acting on the lever A GB. Let these directions
be produeed to meet in a point (0).

Then we may suppose P and W both to act at 0, and
obtain their resultant OB by the parallelogram of forces. It
is evident that this resultant must pass through the falcrum
G, for a single force could under no other circumstances keep
the lever at rest.

Draw Gm parallel to OP. C^, Gw, perpendicular to OP,
W, respectively. Then the sides of the triangle GOm are
evidently proportional to the forces P and TF, and. the reaction
at the fiucrum which will be equal and opposite to OB.

ON MACHINES.

107

rp. n W _ Om _ sin 0(}m _ OGmvl COp _ Cp
inereiore p- ^^- ^^^GOm" OC.COm ""6^*

whieh is the same expression as that obtained in the previous
article.

6. The different kinds of bàlances, or instruments invented
for the purpose of weighing, give us some interesting ex-
amples of tne practical use of the lever.

The common balance is a heavy lever of the first order, from
its extremities A and B two scalea E and F are suspended.

In one of these scales is placed the substance whose weight
it is wished to ascertain, and in the other enough of a pre-
pared series of weights to balance it. The scales must be of
equal weight, and the anns AG, GB, of equal length and
weight. Under these circumstances it will be seen that the
centre of gravity of the whole balance will be a point verti-
cally below G, and therefore, when the scales are empty, or
equally loaded, the beam A GB will rest in a horizontal
position.

There are three reqnisites to a good balance — Horizontality,
StcMUty and JSensibtlity.

Horizontaliiìf is exhibited by the lever or beam resting in a
horizontal pogition, when the scales are empty or equally
loaded. This property may always be secured, as has been
stated above, by making the strings or chains which support
the scales suflSciently long to ensure the centre of gravity of
the whole balance being helow the fìilcrum G.

108

ON MACHINES.

Stahility is exliibited when the balance after being disturbed
quickly returns to ita position of equilibrium.

Sensibility is the property which causes the balance to
indicate a very small difference in the load of the scales by
a large deviation of the beam from its horizontal position.

To fina a measurefor the senaibiUty of a balance.

Let AB be the beam of the balance. Let G, a point below

C, be its centre of gravity. Let 8 be the weight of each scale
together with the tackle suspending it from A or B.

Now let P and W be two unequal weights put in the scale
at A and B respectively, and let A'B', making an angle
with the horizon, be the position which the beam takes in
consequence.

The lever is here kept at rest by three forces {S+P) at
A' {8+W) at B\ and the weight of the beam {w suppose)
acting at O. Therefore, taking moments about C,

{8+P)A'Gcoae + wGCsin0-{8+ W) .ffCcoaB^O;

and slnce (7 is a fixed point this is the only condition of
equilibrium.

And A'C=^B'G=BC; .'. dividing by cos 0,

P.BC+w. GCieLn0'-W.BC=^O;

W-'P BC

tan0 =

' GC

ON MACHINES.

109

Since varies as tan between such limits as will suflSce
for our present purpose, the sensibility will be greatest when
tan^ is greatest; that is, (1) when the. arm [BC) of the
balance is as long as possible, (2) when the weight w of the
beam is as small as possible, and (3) when the distance
between the centre of gravity of the beam and the point of
suspension is also as small as possible.

7. The Common Steelyard is another form of balance.
It is represented in the annexed figure :

Vi/

AB is an iron bar tuming about the edge (7 of a triangular
piece of metal, which works in an aperture of the handle by
which the balance is suspended from a point 0, -

At ^ a hook or a scale is attached for the purpose of re-
ceiying any article whose weight it is wished to ascertain.

P is a moveable weight which can be suspended from any
point between C and É.

Suppose TF to be a body suspended from A, then P must
be moved towards G or towards -B, until the beam AB rests
in a horizontal position.

Let F be this position of P. The bar is so graduated firom
G io B that the reading at F indicates the weight of W,

To graduate the Gommon Steelyard.

Let the weights TFand Pbe removed.
Suppose G^ to be the centre of gravity of the beam,
TF, weight

110 ON MACH]N£S.

The arm CQ being, by construction, considerably heavier
than the rest of the beam, will fall when the machine is in
this unloaded state.

Let the weight P be placed on the other side of the handle
CO at a point 2>, found by trial, such that the beam is kept
in a horìzontal position.

We shall then have

p. CD^ w,. oa (1).

Now let the beam be loaded with P and W as before, then
for equilibrium we must evidently have

p.cF+w^ca=w'.Aa

Substituting from (1),

P{CF+CI))^W.AC;

.Mr=p,5j (2).

Suppose now we put for TF, 1 Ib, 2lbs, 3 Ibs,.,, successively,
we can calculate in each case the corresponding position of F
from equation (2). At the points on the arm so determined
graduations must be made and marked 1, 2, 3 ...

The space between two of these graduations may be sub-
divided into four or more equal parts, and these subdivisions
will indicate with sufficient accuracy the alteration which
must be made in TF to balance P placed at any one of them.
The balance in this way is so graduated that when it is
loaded and Pis placed, by trial, at that ^oint F which makes
the beam remain in the horizontal position, the reading at F
gives the weight of PT.

8. Hie Danish Balance is another form of the common
steelyaxd.

A F G

ON MACHINES, 111

In this case it is the fulcrum which ìs moveable while the
counterpoise remains fixed.

The annexed figure wiU explain ita construction.

To gradìmte thts Balance.

Let the fulcrum G be moved about until the point G is
found on which the unloaded instrument balances. G is the
centre of gravity of the instrument, and its whole weight ( W^)
may be supposed coUected at that point. Now let the balance
be loaded with the weight {W) and let i^be the position of
the fulcrum when the beam is horizontal in this case. The
equation of moments about G will give us

W. AF= W^ . FQ

= W,{AG-AF);

Hence making Tr= llb, 2lbs, 3lbs, ... successively, we shall
be able to mark upon the beam the corresponding positions of
the fulcrum ; and dividing the space between every two such
graduations and a sufficient number of equal parts, we shall,
after setting the instrument, be always able to read off the
weight of the substance suspended at A.

9. The foUowing is an explanation of another balance
which may be termed the Beni Lever Balance.

A GB is a bent lever moveable about a fulcrum C. At the
end A of the arm AG 9, scale-pan is attached. The end B
of the other arm GB is a pointer moving over a graduated
are ED. There is also a knob of metal at B to counterpoise
the scale-pan.

If a weight W be placed in the scale-pan, the pointer B
will move up towards i), and the reading indicated on the
graduated are will give us the weight of Tr .

To graduate this balance.

Let be the angle BGY which the arm GB makes with
the horizontal. Then, the weights and centres of gravity of

112 ON MACHINES.

the arms AC^ GB, being known, and also the weight of the
scale-pan, by taking moments about C we shall obtain an

expression for in terms of these known quantities and W,
Putting, in this expression, llb, 2lbs, Slbs, ... successively
for TF, we shall fina the con*esponding values of 0y and may,
therefore, mark on the are the places at which the pointer
will rest when the scale is so loaaed. Subdividing the space
between every two such divisions into a sufficient number of
equal parts the balance will be ftilly graduated.

This method would prove in practice very inconvenient and
inaccurate. The practical method of effecting the graduation
is to put 1 Ib, 2 Ibs, 3 Ibs, ... in tum in the scale-pan to mark
the place at which the pointer rests.

10. The compound lever is a machine composed of two or
more levers acting upon one another.

The following is an example of such a contrlvance.

CAB is a lever of the 3rd order, united to A'C'B\ a lever
of the Ist order, by a rod B'A\

W
Find the vaine of -^ in the case of equilibrium under the

circumstances shewn in the figure.
If X be the force at B which equilibrates P, we bave

PxAC^XxBC,
and XxA'G'^WxB'C;

OK UACHINES.

113

therefoie, sulistitatmg for X,

FxAGxA'C'=>WxBCxB'C';
W ACxA'C

• P BOxB'C"

jé!

The student may now apply himself to Examinatlon Paper
No. I. at the end of this chapter.

The Wheel and Axle.

11. This machine is shewn In the figure.

It consista of a wheel AB, and a cylinder HS\ both tum-

r5^

ing round the same axis. The pivots C, C\ are placed on
bearings so as to allow of motion round the common axis.

B. S. o

114

ON MACHINES.

The power is applied at the circumference of the wheeì,
often by a cord wrapped round it, as ijii the figure, and the
weight is attached to a cord coiled round the cylinder in the
contrary way, so that when P descends, W ascends, and vice
versa.

The figure below represents a vertical section of the machine
by a piane perpendicular to the axis. The power and weight

are for simplicity's sake supposed to act in the same piane.
is the common ceutre. iJraw through the horizontal
line AB OD.

A and D will evidently be the points where the corda con-
nected with power and weight leave the wheel and cylinder
respectively.

Suppose the figure to represent the machine at one mo-
ment of its action, We might for that moment suppose
the whole machine reduced to a simple lever ABOÌ), of
which is the fulcrum, and A and JD the points of ap-
plication of the power and weight respectively. We should
not by this supposition affect in any way the forces at work.
Hence we see the wheel and axle is an infinite number of
levers, the common fulcrum being O, the arm at which the
weight acts being always equal to the radius {OD) of the
cylinder, and that at which the power acts equal to the
radius ( OA) of the wheel. Hence the wheel arid axle is
sometimes called the Perpetuai Lever.

OK MACHINES.

115

12. To find the moduliLS of the wheel and axle.

Referring to the last figure, it will be evident from what
hàs been already said, that in the case of equilibrium, taking
moments about 0, we must have

Px^O= PTx OD]

WAOR

t •

where R is the radius of the wheel and r the radius of the
cylinder. Which gives us the modulus.

13. The WindlcLss is a modification of the wheel and
axle.

Here the power instead of acting at the circumference
òf a wheel is applied at the extremity A of an arm AB^

as in the figure. It is evident here that if the power be
applied perpendicular to the àrm AB, we shall have in case
of equilibrium

TF__ length of arm AB

F radius of cylinder *

14, The Capstan is another example of the wheel and
axle.

Here the axle or cylinder is vertical, and the power ap-
plied at the end of bars or handspilcea by persons who travel
round as the machine moves so as always to press perpen-
dicularly against the bars.

8—2

116

ON MACHINES.

Here we have

TF__ distance from the hand to the axis of the cy Under
P "" radius of cylinder

15. A combinatlon of wheels and axles may be made in
the manner shewn in the figure.

A, -B, (7, are three wheels, a, i, e, the correspondlng axles.
The power is applied to -4. ^ is connected witn B by means
of an endless strap passing round the axle a and round the

wheel B. This strap communicates motion by means of
friction between its surface and the surface of a and B.
B and G are connected in the same way and the weight W
is attached to the axle o.

OK MACHINES.

117

Now if The the tension of the string between a and -B,
r h and (7,

taking moments ahout the thtee centres, we have, as the
equation of equilibrium for each wheel and axle respectively,

T_ rad, of wheel A
P "" rad. of axle a '

T ^ rad, of w heel B
T rad* of axle h '

Tr^ lrad.of wheel G
T* "" rad, of axle e

Multiplying these equations together,

W prodnct of radii of ali the wheels
F product of radii of ali the axles

which gives ns the modulus òf the combination»

16. It has been said that toothed wheels are sometimes
considered as a separate mechanical power. It ìa better,

perhaps, to consider them as a modification of the wheel and
axle.

US ON MACHINES.

It is evident that the combination described in the pre-
viouB artìcle would not be altered by placing the wheels as
closely as possible together. Let, therefore, two wheels and
axles be placed dose together, and connected by teeth as in
the figure ; if one of them be made to revolve about its axis
by a force P, motion will be communicated to the other by
means of the teeth, and a weight W raised or any resistance
overcome by the motion thus imparted.

Let By r, be the radii of P's wheel and axle respectively,

...• B\ r\ TF's

We shall bave by a process similar to that of the pre-
ceding article,

W_ R r'

if the two axles are equal ; since the teeth are set at eqnal
distances, and their number, therefore, proportional to the
radius of the wheel on which they are set ; the above ex-
pression becomes

W__ number of teeth on PT's wheel
P "" number of teeth on P's wheel

The forni of the teeth is a matter of great importance.
They must be so constructed as to roU upon one another, and
not rwS. We have not only to determme the bestform for
this purpose, but one which will at once produce little rubbing
and admit of being readily cut out in practice. The subject
is very fiilly discussed in Willis's Principlea ofMechanism.

The Pulley.

17. The pulley is a wheel of wood or metal, capable of
turning about an axis through its centre. It is fixed in a
frame-work which is called the hhck or sJieaf. A groove
is cut in the circumference of the wheel so as to retain a rope
passìng over it.

When the block remains stationary the pulley is said to
hejixed, when otherwise, it is called moveable.

ON MACHINES:

119

The annexed figure represents a single fixed pulley.

AB la the fixed block, C the axis round which the wheel
tums, PEFW the string running in the groove over the
wheel.

n I

The power ìs represented as applied by a hand to raise
a weight TF.

As the axis of the pullejr is supposed to be quite smooth,
it is evident that the tension of the string PEFW is the
same throughout. Hence the modulus of this machine is

But though it gives us no mechanical advantage, the con-
trivance is often very convenient for changing the direction
of the force ; for instance, by fixing the puUey in a proper
position, we may raise a weight to any height while we
ourselves are standing on the ground. The effort we must
exert is, however, the same which we should have to exert
if we were lifting the weight without the intervention of
any such contrivance.

The direction in which the power is applied will make no
difference in the equation of equiHbrium P= W\ it will, how-
ever, afiect the strain on the axis, for if 6 be the angle be-

120

ON HACHINES«

tween the dìrections EP and FW^ it is easilj shewn that the
pressure at the axis = 2 TT cos - .

18. The single moveable jmlley.

-4 is a pulley, to the block of which the welght to be
raised is attached.

[fi

]

The rope is made fast to a point O^ and passes linder the
pulley Af and may then, for convenience, he made to pass
over a fixed pulley B, as in the figure, the power being ap-
plied by the hand at P. The pulley A is called a single
moveabte pulley.

To find the modulus of this pulley.

Let F, Ey be the points where the rope meets and leayes
the pulley -4, and Q tnat where it meets tue pulley B.

Let Tj, Tj be the tensions of the stringa CF, OE respec-
tively.

OX MACHINES.

121

The puUey A is kept in equilibrium by the three forces
r„ 2; and W.

Hence, if GÈ he parallel to CF,

Bnt the tension of the string PBOEFC ìs the same through-
out, therefore Tj, T^ each equal P;

.•• 2P= TT,

2»

Hence by this pulley we are able to raìse a wéight of
twice the intensity of the power,

CoK. If the power is not applied in such a way as to make
the cord OE parallel to CF^ let it act in the direction shewn
in the annexed figure^

It will be evident in this case that the portions of the cord
CF^PEwMi take a position equally inclined to the vertical.

Hence, if ^ be the angle between them, resolving the forces
vertically WB shall bava.

2Pcos 5 = TT,
À

122

ON MÀCHIKES.

W .
p = 2C0S-,

which is the modulus in this case.

19. To find the modulus of a system ofpulleys in which each
pulley hangs hy a separate string*

This combination is generally spoken of as the first system
ofpulleys.

Here the last string is attached to a fixed point K^ and
passes under the pulley 0^, the other end being fastened to

jSl

the block of the pulley Oj,, and so with the other strings as
is explained by the figure. The last string passes over a
fixed pulley AB,

To find the relation between the power and weight.

Let 2;, T„ 2; be the tensions of the strings GHK, EFN,
ABGDM.

ON MÀCHINES« 123

Then 2T^^Wy

Whence multlplying and dividing by common factors, we
have

2x2x2?;=» W.
But 2; = P;

If there were four moveable puUeys we should have had

P:=— •
2*'

and similarly if there are n moveable puUeys we should have

W 2"

20. To find the modulus in a system in which the same
string passes round ali the pulleys,

This combination is called the Second System of Pulleys.

Here there are two blocks, in each of which two or more
wheels are fixed. The lower block is moveable, the upper one
fixed. The weight is attached to the lower block, and the
power appHed as in the figure. , The wheels of the pulleys
are of such relative sizes that the different portions of the
string between the pulleys are ali parallel.

It is clear, in this case, that the tension of the string is the
same throughout and equal to P.

Hence, if there be n pulleys altogether, the weight TFwill

OH HÀCHINE8.

l>e upbeld hy n teiiBÌons, and we shall bave in the case of
equìubrìum,

Sometiines the weight of the moTeahlepuIley ìb ao considera
able that it masi he taken into account. When this is the case,
if w be the weight of the lower pulleya and block, by wrìting
W+ u> fot Wm the preceding reault wQ shall make the ueces-

0N MACHINES.

125

sary correction. And in the same way we may allow for the
weiffht of a moveable puUey in any other combination. If there
be a system of moveable puUejs we must add the weight
of each to the tension of the string which supporta it.

21. White's Pulley.

jSl

If in the system of pulleys
we have described in the pre-
ceding article, we suppose, fbr a
moment, ali the pulleys e^ual,
it will be seen, upon consider-
ation, that when W is being
raised by P, the lengths of rope
which pass in the same time
over the wheels of the lower
block are in the arithmetical
series 1, 3, 5, ... while the cor-
responding lengths which pass
over the wheels of the upper
block are in the proportion

ia, 4, O, • • .

We might, therefore, so ar-
range the sizes of the different
wheels, that theportions of rope
which passed over them would
make them ali complete one
revolution in the same time.
The wheels of the upper block
might then be ali glued to-
gether and made to revolve as
though they formed one piece on a common axis.

This is the principle of White^s mdley. The wheels are
cut out on the faces of two solid blocks, the upper one of
which is fixed and the lower moveable as in the figure. The
modulus of this pulley will be the same as that of the second
system.

22. To find the modulus of a system of pulleys in which ali
the strings are attached to the wetght»

This combination is called the Third System of Pulleys,

126

OS MACHINES.

jH.

The figure will explain ìtself.

Let Tj, T^ Tg ... be the tensions
of the strings.

We shall clearly have 1\ = P.

The second string, whose ten-
sion is Tj, has to support two
tensions each equal to P.

Hence T^ = 2P.

And similari^

!r3 = 2?;=-2«p.

And j; = 2?; = 2'P.

And 2;«2r^, = 2"-^'P.

Hence

= P+2P+2"P+...+2"-'P
= (2"-l)P;

TF_ 2*--1

23. There are many more combinations of puUeys, indeed
they are almost infinite in number. The student will not
fina any diflSculty in dedacing the relation between the
power and the weight in the case of equilibrium, in any
system that may be propósed.

A combination of pulleys is sometimes called a Barton^ or
a Spanùh Barton.

It has already been noticed that the weights of the pulleys
sometimes are so large as to make it necessary to take them
into accomit, and a method of introducing them into our
calculations has been pointed out. There are other and more
important causes which affect the accuracy of our results.
The thickness ofthe ropea is sometimes a very sensible quantity
in proportion to the diameter of the puUey ; where this is so,
half the thickness of the rope must be added to the diameter
of the wheel of the puUey, in ali expressions where that
diameter makes its appearance; and a similar precaution

ON MACHINES. 127

must be taken in proUems on the wheel and axle. Another
Bonrce of error is tne friction of the axù ; a method of calcu-
lating the amount of this will be given hereafter. But the
most.firuitful source of inaccuracy is the rigidity of the ropes
employed. When the ropes are new and the puUeys of small
diameter, a considerable portion of the power is expended in *
bendin^ them. Tables nave been constructed for shewing
what allowance must be made in our results on this account.
But the practical experience of the engineer will generally be
a sufficient guide on the subject.

I%e Inclined Flane*

24. The Inclined Piane is merely a piane inclined at an
angle to the piane of the horizon.

The angle which it makes with the piane of the horizon is
called its inclinatton.

The inclined piane is most frequently used to enable us to
raise a body by a force less than the weight of the body.

The general problem ìb to find what force P, acting in a
given direction, will support a given weight W, refeting upon
a piane of given inclination ; or, to find the modulus of the
machine in this case.

Let a be the inclination of the piane to the horizon,
li the reaction of the piane,
W the weight of the body,
JP the force acting at an angle e to the horizon.

128 ON HÀCHINBS.

The body is evidentlv kept at rest by the three forces P,
Bf W, and resolving along and perpendicular to the piane
we bave

Pcos€- Trsina = (1),

Psine- Trcosa + 5 = (2);

firom (1) we bave

W COS€

P Sina
whicb gives ns the modulus.

And from (2) we may find R the pressure on the piane,
which will be found to be

_ cos (g + e)
"" cos e

If P should act parallel to the piane, by pntting 6 = the
above expression becomes

W 1

Sina

If P should be horizontal we must put € equal to — a, and
the expression becomes

-rp = cot a.

The problems connected with this machine involve no new
principle; the manner of treating them has already been
explamed in preceding chapters.

The Screw.

25. This mechanical power is a combination of the lever
and inclined piane. It consists of a cylinder with a con-
tinuous projecting thread wound round its surface. This
cylinder works into a block in which there is a groove cor-
responding to the projecting thread ; or the cylinder may bave
a groove and the block a projecting thread.

OK MACHINES.

12»

The thread or groove is of such a character as to be inclined
at eveiy point at the same angle to the axis of the cjlinder.

The thread of the screw may be conceived to be generated
by wrapping an inclined piane round a cylinder. Cut out
a piece of paper in the shape of the right-angled triangle
AÈC, place the side BC on a pencil or some other cylindrical

surface so as to be parallel to the axis of the cylinder ; then
wrap the p^aper round the cylinder and AB will mark out a
continuous descending path, which will be the thread of
a screw.

The screw ìs generally worked by means of a bar or arni

inserted into the head of the cylinder perpendicular to ìts
axis. One revolution of the cylinder will cause the screw to

B. S. Q

130 ON MACHINES.

be pushed through the block or nut a distance equal to that
between two tbreads of the screw.

To find the modulos in the case of the screw.

Suppose the screw to be vertical and to be used for raising
a weignt TF, the power P bein^ applied at the extremitj of
an arm, and at right angles to it, the arm and the direction
of P both being in a horizontal piane, as in the figure.

The machine is evidently kept in equilibrium by the force
W acting downwards, the force P, and the reaction of the
thread at eveiy point in contact with the groove represented
by the forces ^, -B, ... in the figure.

Let now a portion of the thread be conceived to be
unwrapped from the cylinder, let AB, the hypothenuse of the

right-angled triangle, ABC, be such a portion ; BC being the
distance between the threads, and Au the circumference of
the cylinder.

Consider the equilibrium of a point 0^ on the thread.

Let B be the reaction at that point perpendicular to the
thread,

w^ the portion of W which is supported at 0^,

^1 the portion of P which is expended at the same point.

Hence if a be the angle BA C, resolving along the line AB,

p^ cos a *= tt?j sin «.

ON MACHINES. 131

Therefore -^ = tan a

"AC

vertlcal distance between two threads
circumference of cylinder

And similarly, ìiw^w^...pjp^...hQ the quantities corre-
sponding to w^^ ai points Oj u, . . • respectively , we have

vertical distance between two threads _Pt^Pi^Ps^jgr
circumference of cylinder "" ««^i ~ 'W'a ~ w^a ~

^ Pt±p^±Ps±&c^
w^ + w^ + w^ + &c. '
and w^ + w^ + w^ + &c. = W.

And since Oj, ^j, ... ali act at an arm r (if r be the radius
of the cylinder), we must have, if a be the arm at which
Pacts,

( A + ft + &c.) r = P.a;

P vertical distance between two threads

^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

- (circumference of cylinder]

_^ vertical distance between two threads
circumference of circle described by P'

which gives us the modulus.

The Wedge.

26. The wedge is a solid triangular prism formed of some
hard material, as ìron. Its most common use is to separate the
parts of a body, which have from the nature of the material
a tendency to approach one another. This is done by intro-
ducing the edge of the wedge between them, and then driving
it forwards with the blows of a hammer.

9—2

132

ON MACHINES.

The figure represents the section of a wedge by a piane
perpendicular to ita edge. This section is an isosceles
triangle.

To find the modulus.

For simplicity's sake we suppose the wedge quite hard and
smoòth, and to be inserted between two symmetrically situated
obstacles.

Let Jlf be the poìnt of contact of one obstacle with the
wedge. Since the wedge is smooth, its action must be per-
pendicular to its surface. Let this normal action be B. The

action of the obstacle will take place in the direction in which

W
M tends to move. We may cali this resistance — (since

there will be the same resistance on the other side of the
wedge, and the whole resistance will correspond to what we
bave in other machines called W). Let % be the angle which

the direction of — makes with R.

ON MACHINES.

Thea since the normal force B must balance the force

Ì33
W

we bare

5 C08 t = —

Jé

(1).

Again, if P be the force with which the wedee is driven
downwards, since half this force will be expanded on each
side of the wedge, resolvinff vertically we have, if a be the
angle at the edge of the 'wedge,

P P . a
- = 5sm-

Substituting for B from (1),

• ■

P W ^^1
2 2 cos e '

W cos t

P TU'
sin-
2

which gives us the modulus.

(2).

Machines mth Frtction,
27. Sometimes the fulcrum of the lever is formed by

a circular hoop rigidly attached to the lever riding upon
a cjlindrical axis. The necessary correction on account of

134 OK MACHINES.

frictìon in this case maj be obtained in the foUowing
manner.

Let the figure represent a lever AB rigidly attached at
the point m to the outer circle mba^ and let noe be a section
of the cjlìndrical axis.

These two circles, being nearly equal, may be considered
to bave a common centro 0.

When AB is horizontal, the directions of P and W are
sapposed to be perpendicular to the length of the lever, and
m to be vertically over 0.

Let the figure represent the limìting position of eqtiilibrium,
i.e. the position as far as possible removed firom the norizontal
one consistently with equilibrium.

Let n be the point of contact of the two circles in this
case. It is evident that the forces must be in equilibrium
about the point n.

Draw OnB to represent the normal pressure B at n.

Braw ufiB to represent the friction oB lu, being the coeffi-
cientoffriction).

Then resolving the forces along the normal and tangent
at n, and taking moments about the same point we bave,
if angle n 0971 = ^,

(P+ W) cos5-jB = (1),

(P+Tr)sin5-yx5 = (2),

and TFx nw? — Px wp = (3),

where nw and np are the perpendiculars from n on the direc-
tions of Pand fr'respectively,

But if ToB^ &, mA = a, and nO=^r, we bave

nw^h + rAnOy
Tip =5 a — r sin ^»

ON MACHINES. 135

Therefore (3) becomes

Tr(5 + r. sin d)-P(tt-r. sin ^ = (4).

(1), (2), and (4] will suffice to determine the ralue of Pin
the case of equilibrium.

From (1) and (2) we hav»

tasid^fjL (5),

^wliicli detennines the point of contact.
From (4),

From (5),

Therefore

sm ^ = — ^-— -: .

(1+/*')*

fi Pa-W.h

or Pa=W.l+ , ^ , . .(P+HO.r;

*+ /^

i.e. P=Tr. ^IJJLL

1- /^ I

J + r . sin e
a — r. sin 6*

= TF

where for the sake of convenience we pnt ^ = tan€. In
this case e is equal to 0^ and therefore the result might be
obtained directly from equation (4). If the friction acted in
the contrarj waj, we could immediatelj find the limiting

136

:0N MACHINES.

value of P in tliis case hj changing /i into — yx, or sin € into
— sin € in the above formula. Hence if P^, jPj be the two
limiting values of P, we bave

^ a— r.sine
* a + r.sme

where

sin€ =

(!+/.»)*'

and any vaine of P between P^ and P, will produce equi-
librium.

28. In the case of the wheel and aosle the result will be
precisely the same as the above, a and b being the radii of

the wheel and axle respectively. The figure shews, suffi-
ciently, the modification of the problem. The friction arising
from the axis on which the wheel of a pulley runs will also
be treated in the same w«y.

29. The case of the inclined piane with firiction offers no
difficulty.

Let TT be a weight restìng upon an inclined piane whose
inclination is a.

ON MACHINES.

137

Let it be acted upon bj a force JP making an angle e with
the piane.

Let TTbe on the point of sHpping dotm the piane, and let
P^ be the vaine of P m this case ; draw R the normal reaction

and the force of fnction (jiB) as in the figure. Besolving
along and perpendicular to the piane, onr equations of equi-
librinm are :

PjCOB€ + /xuB— TTsin a =
^ + Pi sin e- TTcos a =

0),

(2).

Substituting in (1) the vaine of jB from (2),

PjCOse + /A (TT. cosa — PjSine) — TFsina = 0,

p _ "FT. sin a — /aTTcos a
* "" coB € — /Lt sin €

If TF be on the point of moving up the piane, and P, be
the vaine of P, in this case

P — ^sing + AtTrcosg
a resolt which ìs obtained hj merelj changing the sign

of /Lt.

138

ON MACHINES.

And tliere will be equilibrium for any vaine of P between
P, and P,.

30. We bave next to consider the effect of frictìon on the
screw.

Lety in tbe figure, AB be a portion of tbe thread of the

screwnnwrapped, BO being the distance between two threads,
and A G the circumference of the cy linder.

If the screw be on thepoint of moving npwards, the frictìon
will act downwards along the line -4P, since AB will coincide
with the tangent at the point 0. of the thread, and using the
sanie notation and snpposin^ the screw to be worked in the
same way as in Art. 25, resolvin^ along and perpendicular to
the line AB, we bave for the equilibrium of the point 0^

p^ cos a *- t«?i sin a — fiB ^ 0,

2>i sin a + tt?j cos a — P = 0.

Whence elimìnating P,

Pi _ sina + yLtcosa
t(?j"cosa— /tsina*

And if ^,, «?, be the quantities corresponding to p^, w^
for anj other point in the thread 0^, we snall bave exactly

the same expression for " , and similarly for every ofher point.

ON MACHINES. 139

cosa — /*sma w^^ w^ w^

and «?i + «?g + w?, + &c. = W.

And if Pj "be the amount of the power applied at the end of
the arm when the screw is just moving up, we shall have, as
in Art. 25,

Hence, substituting in the above expression,

P^ a ^BÌna + fi. cos a
TT V" cos a— /A . sin a *

p_^ sin g + /i . cos g .^^7-

or xl — • ; • rr •

* a cosg^/i.sina

And if Pj be the vaine of P, when the screw is on the point
of descending,

P — ^ SIP tt — M » CQS g 1--^
^ a cosa + /;<fr.sma

Any vaine of P between Pj and P, will produce equi-
librium.

31. Friction might be introduced into the expression
already obtained for the equilibrium of the wed^e without
much difficulty, but the results obtained for this machine
are from various causes of so little practical use that it seems
unnecessary to dwell longer upon them«

140 ON MACHINES.

Examinatùm on Chapter VII.

Paper I.

1. Define the term Machine. Enumerate the machines
known as the six mechantcal powera, What do you mean by
the moduhs of a machine ? When is a machine said to work
at a mechantcal advantage f

2. Define a simple or mathematica! lever. Into what
three orders are levers divided? Give an.example of each
order.

3. Enunciate the properbr which is the condition of equi-
librium of the lever; and fina this condition, by a geometrica!
method, in the cas^ of two forces not pakllel «^ting ou a
straight lever.

4. A weight of 1 cwt. is required to be raised with a
straight !ever of the first order, 10 ft. long, worked by a
power of 8 Ibs.; find the position of the folcrum.

6. A simple lever of the first order, the arms of which
are inclined to one auother, without weight, has two weights,
P and W^ suspended £rom its extremities ; the lever rests in
such a position that the string of P makes an angle of 60^,
and that of W an angle of 45^ with the lever. The arms
of the lever are respectively 3 and 5 feet long ; find the ratio
of Pand W.

6. The arms of a bent lever are inclined to one another
at an angle of 150^, from their extremities weights of 7 and
6 Ibs. respectively are suspended. The lengths of the arms
are 3 and 5 feet. Find the inclination of each arm to the
horizon when there is equilibrium.

7. Describe the common balance. What is meant by
the aensihility of the balance? Obtain an expression for
measuring the sensibility, and point out what is necessary in
the construction of the balance to secure this property.

ON MACHINES. 141

8. A common balance has ita arms of unequal length;
a body when placed in one scale appears to weigh 4lbs.,
when placed in the other, 16 Ibs. What is tìie trae weiffht
of the body?

9. Shew bow to graduate the common steelyard.

10. Shew how to graduate the Danish Balance.

11. What is a compound lever? Three levers of the
first order are so arranged that the first applies the power to
the second, and the second to the third, the powers and weights
are ali perpendicular to the lengths of the levers. Find the
modulus.

12. A bent lever of uniform thìckness, suspended by its
fulcrum, rests with its shorter arm horizontid; but if the
length of thìs arm be doubled, the other arm would be hori-
zontal. Compare the lengths of the arms, and find their
inclination.

Paper II.

1. Find the modulus of the wheel and axle. Explain
why this machine has been called the perpetuai lever.

2. If the rope used be of such thickness that it becomes
necessary to take it into account, what will be the expression
for the relation between P and W?

3. A weight of 1000 Ibs. is sustained by a rope of 2 inches
diameter ; the diameter of the wheel is 1 J feet, and that of
the axle 6 inches. What must be the magnitudo of the
power to produce equilibrium?

4. There is a combination of wheels and axles, connected
together by straps passing roimd an axle and the next wheel,
the ratios of the radii or the wheels and axles are respec-
tively 2 : 1, 4 : 1, 8 : 1, &c. The modulus of the machine
equals w. Find tìie number of wheels.

142 ON MACHINES.

5. Find the modulus of the single moveable pullej;
the strìngs not being parallel to one another.

6. At what ande must the strìngs be incllned, in the
single moveable piuley, in order that P may equal W?

7. Descrìbe the arrangement of the^r^^ system ofjndleySf
and find the modnlus.

How manj pulleys must there be in such a system that
1 Ib. may support 128 Ibs.?

In the same system shew that if P and W and the
weight of each pulley be ali alike, there will be equili-
brìum.

8. Find the modulus of the second system of puUeys,
and descrìbe the prìnciple of White's pulleys.

What must be the magnitudo of P in the second system
in order that it may just balance the weight of the pulleys ?

9. Find the modulus of the third system of pulleys.

There are 6 pulleys in such a system ; find the ratio of the
weight of each pulley to W, in order that the weights of the
pulleys may just support W.

10. A weight Ì8 supported on a smooth inclined piane by
a force acting along the piane ; shew that P is to TF as the
height of the piane is to the length.

11. A weight W is just supported on an inclined piane
by a force P, acting by means of a wheel and axle placed at
the top so that the string attached to the weight is parallel
to the piane. Given the radii of the wheel and axle, find
the incunation of the piane.

12. Find the modulus of the screw.

13. What weight can be sustained by means of a power
of llb. acting on a screw, the distance between two contiguous
threads of which is 1 inch, and the power being applied at

PBOBLEMS, 143

the extremitj of an arm 3 jis. dìstant firom the axis of the
screw?

14. Find the modulus of the wheel and axle, taking into
accoTmt the friction of the axis.

15. The inclmatlon of a piane is known, and ita rough-
ness ìs such that a body will just rest upon it ; find the least
force, acting along the piane, necessary to draw it up.

16. The sliding weight of the common steelyard is 9 Ibs.
The zero-point of graduation is ^ an inch from its fdlcrum
on the longer arm, and the whole beam will balance on a
point 3 ìnches from the fulcrum on the shorter arm ; what is
the weight of the beam?

17. What force must be exerted to sustain a ton weight
on a screw, the thread of which makes 150 tmus in the course
of 12 inches, the length of the arm being 8 feet?

Problems,

In solving statical problems the student will do well to bear
in mind the foUowing rules :

1. Draw the figure, representing ali the forces by straight
lines and arrows, shewing the directions in which they act.

2. Write down ali the equations of equilibrium. Choose
the directions in which you resolve your forces, and the point
about which jrou take your moments, so as to make your
equations as simple as possible.

3. Count the unknown quantities in the statical equations,
and then add as many more equations deduced from the geo-
metry of the figure as will make the whole number of equa-
tions equal to the whole number of unknown quantities
involVea.

4. If the problem involve more bodies than one, the
actions and reactions between them and the forces arising
from firiction must be represented, and the equilibrium of each
body considered separately.

144 PBOfiLEMS.

1. A particle whose weìght is 4lbs. is suspended freelj
from a point, and acted upon by a borizontal force equal to
3lbB.; nnd the direction and magnitude of the force neces-
sari to keep ìt from moving.

2. A smooth rin^ is suspended from a point. A string
passes through the ring and is pulled at either end with a
torce of 5 Ibs. in such a manner that the angle between each
portion of the string so pulled and the string which suspends
the ring is 30^ Find the strain upon the point of suspen-
sion.

3. The resultant of two forces represented by 1 and 2
makes an angle of 150^ with the greater force* Find the
magnitude of the resultant.

4. Three forces P, P, and mP keep a particle at rest ;
determine the angle between the equal forces.

5. A particle placed on a smooth table ìs pulled by three
strings with forces equal to 3-F, 4F, 6jFrespectively, and kept
at rest. Find the angle between the first two strings.

6. Two forces acting at right angles have a resultant
which is doublé the smaller force ; find its direction.

7. A particle placed on a smooth table is pulled with a
force of 5 Ibs. by a string which makes an angle of 45^ with
the piane of the table. What ìs the force which tends to
make the particle move along the table?

8. A pillar is to be pulled down by a rope of given
length. At what point in the pillar must ìt be tied in order
that the least force may be sufficient for the purpose?

9. Three forces represented by 2, 4, and 9 pounds re-
spectively act upon a particle. Shew that they cannot pro-
dxice equilibrium.

10. ABy AGy two chords of a circle, represent two forces
in mamitude and direction, AB being given, find the posi-
tion ^ AG that the resultant may be a maximum.

PBOBLEMS. :]45

li. Two heavj ]^article8 acQ united hy a striag, a&d lest
upoB a smooth. vectical circle^ find thQ poaition oi equiU-

12. Three smooth tacks are arranged on a vertical wall so
a9 to fona an ieoaceles triangle wIxosq vertical aa^le is 120^,
the base of the triangle being horizqntal. A string with a
weight (P) fastened to each end ìa then laid over th^. Fisid
the strain on each tack. •

13. A string has one extremity festened to a point A ; it
then passes through a smooth ring which supports a weight,
and tne other extremitj* is fastened to a point B in the same
liorizontal line with A. Given the length of the string and
the distance between A and J9, find the strain on A and R

14. In the last example, suppose the line AB not hori-
zontaT. Giren its length and tne angle it naakes with the
horìzon, and the length of the string, find the poaition of
equilibrium.

15. A string is fastened at one extremity to a point A,
and then passes over a smooth peg B, a weight P being
attached to the other extremity. A weight W is suspendea
from the point C in the string ACBP. A and B are in the
same horizontal line.

Given AB —AC^ a, find the poaition of equiliWum.

16. A weight Wìs suspendcd by a string^^CTTi^oin the
point A. At the point G a horiz;ontal force F acts on the
string, what must be the magnitude of F that the angle ACF
may be 60\

17* Two men standing on opposite aides of a post pulì at
it by means of two ropes of the lengtha ?, ?, respectively,
fastened to the same pomt in its length. What must be the
raiio of the strength of their puUs, in order that their efforts
to draw the post out of the perpendicular may just neujtralize
one anotherr

18. Fi ve forces act upon a particle. Their magnitudes
are represented respeatively by tne figures 1, 2, 3, 4, 5, their
R, S. 10

146 PBOBLEMS.

directicms makc the angles 60^, 120^ 180*, 240", 300*, witt a
'fixed line. Fìnd the magnitnde and direction of a force
which will just connterbalance the tendency of the particle to
move.

19* A weight W is snstained on a Bmooth inclined piane

W
hj 8 fprces each eqnal to — , one acting vertìcally npwards,

another horizontallV, and a third parallel to the piane ; find
the inclination of tne plane.^

20. Two given weights P and Q rest upon two smooth
inclined planes, and are attached together by means of a
Btring passing o\«er the common vertex of the planes. Find
thè ratio of the inclinations of the planes.

21. Shew that if a system of forces be represented in
magnitude and direction by the sides of a polygon taken in
order, the moment of the forces about any point may be
xepresented by twice the area of the polygon.

22. A circular are without weight rests with ìts convexity
downwards on a horizontal table. Two weights P and Q are
suspended fìrom its two extremities. Determine the position
òf equilibrium.

23* Two parallel forces acting upon a rigid line, in oppo-
site directions, are represented in magnitude by the figures
3 and 6, and their diréctions are 12 inches apart; fina the
magnitude and position of a third force necessaiy to keep
them in equilibriim.

24. A given uniformbeam rests with one end in a smooth
hemispherical bowl, and the upper end against a smooth verr
tical Wall. Find the position ot equUibrium.

25. A piece of wire is formed into a triangle ; find the
distance of the centre of gravity from each of the sides ; and
shew that if a;, y, 2; be the 3 distances, and r the radius of the
inscribed circle, then

4a?y« — r* (a; +y + ^) — r' s= 0.

PBOBLEMS. HT

' 26. The sum óf the sqùares of the' thrèò sides óf a
trìangle equals three times the sum of the squares of the linea *
drawD from the céntre of gravity to the three angular points
q£ the trìangle,

•

27. If the centre of gravity of a fonr-sided figure coincide
with òné of ita togular points, shew thàt the distances of . this
point and the opposite angular point from the line joining thè
other two angular points are as. 1 to 2,

' 28, Two given heavy particles are connected together "by
an inflexible rod without weight and of given length, and
placed in a smooth hemispherical bowl ; find the position of
equilibrium.

29. Two small rings. slide on a smooth vertical circle; a
^bring passes through both rings and has 3 equal weights
àttached to it, one at each end, and one between the rings;
find the position of the rings in the case of equilibrium.

30. A uniform heavy rod of given length is to be sup-
ported in a given position with its upper end resting agàinst
a smooth vertical wall by means of a string àttached to the
lower end of the rod and a point in the wall. Find the point
in the wall to which it must be àttached.

31. Ab is a rod capable of tumihg freely about its ex-
tremity A, which is fixed ; CD is another rod equal to 2-4J5,
and àttached at its middle point to the extremity B of the
former, so as to tum freely about this point. A given force
acts at G in the direction CA. Find the force which must
be applied at -D in order to produce equilibrium,

32. A given rod rests with its upper end against a
smooth vertical wall and the lower end suspended by a given
string fastened to a point in the wall. Find the position of
equilibrium.

33. A uniform beam AB moveable about a hinge at A
in a vertical piane leans upon a prop CI) in the same vertical
piane ; determine the strain upon the prop.

tlS PBOtBLXMS.

34» A rigìd iród withotit weight lies with one end agàìnst
the smooth mtemal mit&ce^ and the otfaer end pixyjectiiig
bejond the TÌin of a conical ^ell whose axis is verticaL A
weight (W) Ì8 suspended from the projecting end òf tìie.
beam ; find the position of equilibrium of the rod.

35. A he^nii^hete ié pkeed iipon a irough incKned piane ;
^ìvtn die eoeffieteait of "friction betw^n the snrfaoeB ; find the
limiting position of equilflbrinm.

86. A string fastened to a point in a lon^ vertical wall
is wrapped round a ball which is then allowed to bang down
agamst the wall. Dotermine the limiting position of equi-
librium.

37. A.^iniform l)eam xests wIth its middle point upon a
rough veriieal circle ; find the greatest weight that can be
suspended from one end of the beam without its sliding off
the circle.

38. A weight is suspended firom a point in thè circum-
ference of a hoop. The hoop is hung on a rough peg, and
rests so as to be on the popt of slipping round on the peg.
Mnd the position of equilibrium.

. 39. Two circukur discs hang in a vertical piane bj means
of a string which passes over a smooth peg ; the two end» of
ihe string being attached to the (rwo centres of the disca
lespectively. The circumfereaces of the discs aie «nooth^ and
are placed in contact so as to keep the Bjstan in equìiibrium.
Shew that the portbns of the stringa on either «ide of the
peg are to one another as the squares of the radii of the discs.

40« Two eqoal and uniform beams AC, B€ are oon-
ftected bj a dmooth hin^ at (7, and are placed upright on a
horizontal piane with the ends A and B resting on the piane ;
if )3 be the greatest value of the angle A GB which is con-
sifltent with equilibrium, «hew that the ooeffieìent of frietion

= itan-.

PfiOBLEHS. 1^

41. 1£ A home exert a tractiom <tf IMlbs^, wfaat weà^Jit
wiU be pulì up a kill whkk has « rise <3f ^ ia 100 ; sup^oaing
iìkù «oeffioient of fiàetioin to tie ^?

42. Two roiJ^Ii bodies test on an indined piane and ate
connected hy a string which is parallel to the piane. . If the
coefficient of frietion be not the same for both, find the
greatest inclination of the piane consistent with equilìbriom.

43. A triangular wedge resta upon a rough horizontal
table. The side which forms an inclined piane is smooth,
and upon it a heavy particle is supported by a force making
an angle with the inclined piane. Find the whole pressure
on the horizontal table.

44. A uniform beam rests against a peg at the focus of a

?arabola, its lower extremitj being supported by the curve,
'he axis of the parabola is vertical and its sur&ce smooth ;
determine the ìhclination of the rod to the horizon.

45. In the system of pulleys where each string is attached
to the weight, if one stnng be nailed to the block through
which it passes, shew that the p»ower may be increased up to
a certain extent without producine motion. If there be three
pulleys and the action of the middle one be checked in the
manner described, find the tension of each string for given
values of P and W.

46. If the axis about which a wheel and axle tums coin-
cide with the axis of the axle but not with that of the wheel ;
find the greatest and least ratios of the power and weight
tìecessary for equilibrium, neglecting the weight of the ma-
chine.

47. The arms of a balance are unequal, and one of the
scales is loaded. A body whose trae weight is Plbs. ap-
pears to weigh T^lbs. when placed in one scale, and TT'lbs.
when placed in the other ; nnd the ratio between the arms,
and the weight with which the scale is loaded.

ELEMENTARY MATHEMATICAL WORKS.

IL TEIGONOMETEY.

INTRODUCTION to PLANE TRIGONOMETRY. For the
use of Schoola. By J. C. SLNaWBALL, H.A. •ocond Edition.
(1847). 8vo. 5*.

PLANE TRIGONOMETRY. For Schools and CoUegea. By
I. TODHUNTEB, M,A «71 pp. (1859^. Crow» 8w. j*.

SPHERICAL TRIGONOMETRY. For COLLEGES and

SCHOOLSi By I. TODHUNTER, M.A iia pp. (1859). Crown 8vo.
4«. 6d.

PLANE TRIGONOMETRY. Wiih a numerous CoUection of
SxMiiplM. BstB. D* BIBASlifi?» M.A. 106 pp. (1858}. Caorwn Svow

PLANE and SPHERICAL TRIGONOMETRY. With the

Conatructìon and uae o£ Tables of. Logarithivs. By J. G. SNOWBALI^
M.A. Nihth EditìoD, 240 pp. (1857}. Crown 8yow 7«. 6d.

m. MEGHA10C9 AND HTDBOSTATICS.

ELEMENTARY TREATISE on MECHANiIGS. With a

CoUection of Examples. By S. PARKINSON, B.D. Second Edition.
Jn tke J^reu,

ELEMENTARY COURSE of MECHANICS andl HYDRO-

STATICS. By J, a SNOWBALI^ M.A. Fonrth EdiUe». iM pp.
(185 1). Crown 8 vo^ 5*.

ELEMENTARY HYDROSTATICS. With numerous Ex-
amples and Solutions. By J. B. PHEAB^ H.A« Second Edition*
156 pp. (1857). Crown Sto. 59. 6d,

Macmillan aih) Co., Cambridge and London.

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Full text of "Elementary statics, ed. by E. Sturges"

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